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Learning Objectives:

Define charge and Coulomb's law.
Explain the force of attraction between two charged bodies as a vector quantity.

Electric charge (q), is the extent to which a matter has more or fewer electrons compared to protons.
Electrostatic force that one charge, $q_{1}$, exerts on another, $q_{2}$, has a magnitude given by:

\[F = \dfrac{k q_{1}q_{2}}{r^{2}}\]

where r is the distance between $q_1$ and $q_2$.
The direction of this force is either toward or away from the other charge depending on whether it is attraction or repulsion.
In the above equation, k is called Coulomb’s constant and its value is given by: \[k = \dfrac{1}{4 \pi \epsilon_0} = 9.0 \times 10^9 N m^2/C^2\] Coulomb’s law in the vector form can be expressed as: \[\vec F_1=k\frac{q_1q_2}{r^2}\hat r_{21}\] where, $\hat r_{21}$ is a unit vector which is pointing from the other charge to the charge itself.

Solved Example: 18-1-01

The electrical charge associated with one electron, in Coulombs, is:

A. 1

B. -1

C. 6.25 $\times 10^{18}$

D. -1.60 $\times 10^{-19}$

The elementary charge, usually denoted as e or sometimes q, is the electric charge carried by a single proton, or equivalently, the magnitude of the electric charge carried by a single electron, which has charge -e. This value is constant and equals to (-)1.6$ \times 10^{-19}$ Coulombs.

Correct Answer: D

Solved Example: 18-1-02

Two identical point charges with mass $m_1$ = $m_2$ = 0.60 grams are suspended in air with strings of length = 250mm. The total angle between them is 15$^\circ$. The charge each of them carrying is:

18-1-02

A. 19.15 nC

B. 27.45 nC

C. 31.87 nC

D. 12.11 nC

GD and T

Let tension in each string be T.
Equating horizontal components $T\sin 7.5^{\circ }=k\dfrac {q_{1}q_{2}}{r^{2}}$
Equating vertical components $T\cos 7.5^{\circ }=mg$
Dividing above two equations, \[\tan 7.5^{\circ }=k\dfrac {q^{2}}{r^{2}mg}\] \[q^{2}=\dfrac {r^{2}mg\tan 7.5^{\circ }}{k}\] where, \[r=2\times 250\times \sin 7.5^{\circ } =65.26\ \mathrm{mm}\] \[q=1.915\times 10^{-8}C=19.15\ \mathrm{nC}\]

Correct Answer: A

Solved Example: 18-1-03

Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the top right and bottom left corners are +3.0 $\times$ 10$^{-6}$ C. The charges in the other two corners are -3.0 $\times$ 10$^{-6}$ C. What is the net force exerted on the charge in the top right corner by the other three charges?

A. 68.7 N

B. 72.9 N

C. 88.2 N

D. 118.0 N

GD and T

\[F = \dfrac{kq_1q_2}{r^2}\] \begin{align*} F_{14} = F_{12} &= \dfrac{(9\times 10^{9})(3\times 10^{-6})(3\times 10^{-6})}{(0.025)^2} = 129.5\ N \end{align*}

(F$_{12}$ towards q$_2$ and F$_{14}$ towards q$_4$)

\begin{align*} F_{13} &= \dfrac{(9\times 10^{9})(3\times 10^{-6})(3\times 10^{-6})}{(0.025\sqrt{2})^2}\\ &= 64.7\ N\ (\mathrm{away\ from}\ q_3\ \mathrm{as\ shown}) \end{align*}

FBD

\begin{align*} \Sigma F_x &= -129.5 + 64.7 \cos 45^\circ = -83.75\ N \end{align*} \begin{align*} \Sigma F_y &= -129.5 + 64.7 \sin 45^\circ = -83.75\ N \end{align*} \begin{align*} F_R &= \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2}\\&= \sqrt{(83.75)^2 + (83.75)^2}\\&= 118.43\ N \end{align*}

Correct Answer: D

Solved Example: 18-1-04

Three charges 4q, Q and q are in a straight line in the position of 0, $\dfrac{l}{2}$ and l respectively. The resultant force on q will be zero, if Q=? (RSMSSB Lab Asst. 2018)

18-1-04

A. -q

B. -2q

C. -q$^2$

D. 4q

18-1-04

The force between 4q and q ${{F}_{1}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\cdot \dfrac{4q\times q}{{{l}^{2}}}$
The force between Q and q ${{F}_{2}}=\dfrac{1}{4\pi {{\varepsilon}_{0}}}\cdot \dfrac{Q\times q}{{{\left(\dfrac{l}{2}\right)}^{2}}}$
Since, the resultant force must be zero, \begin{align*} {{F}_{1}}+{{F}_{2}}&=0\\ \dfrac{4{{q}^{2}}}{{{l}^{2}}}&=-\dfrac{4Qq}{{{l}^{2}}}\\ Q&=-q \end{align*}

Correct Answer: A

Solved Example: 18-1-05

Two charges of + 4 $\mu$C and -16 $\mu$C are separated from each other by a distance of 0.6 m. At what distance should a third charge of + 6 $\mu$C be placed from + 4 $\mu$C so that no force exerts on it will be zero? (DSSSB JE EE Oct 2019)

A. 0.4 m

B. 0.6 m

C. 1.2 m

D. 0.3 m

Correct Answer: B

Solved Example: 18-1-06

The force between two charges $Q_1$ and $Q_2$ is given by: (NLC GET ECE Nov 2020)

A. Newton's law

B. Coulomb's law

C. Faraday's law

D. Ampere's law

Correct Answer: B

Solved Example: 18-1-07

The force between two charges is 200 N. If the distance between the charges is doubled, the force will be _______. (DFCCIL Jr. Executive S&T Sept 2021)

A. 400 N

B. 100 N

C. 50 N

D. 200 N

Correct Answer: C

Learning Objectives:

Understand the meaning and significance of electric potential.
Use the electric potential to calculate the electric field.
Use integration to determine electric potential difference between two points on a line, given electric field strength as a function of position along that line.
Calculate how much work is required to move a test charge from one location to another in the field of fixed point charges.
Express in equation form the energy stored in a capacitor.

Electric Field:

Electric field is defined as the electric force per unit charge. Electric field is also known as electrostatic field intensity.

Electric field from a point charge:

Electric Field from a Point Charge

The electric field a distance r away from a point charge q is given by:

\[E = \dfrac{k q}{r^2}\]

The electric field from a positive charge points away from the charge; the electric field from a negative charge points toward the charge. Like the electric force, the electric field E is a vector. If the electric field at a particular point is known, the force a charge q experiences when it is placed at that point is given by :

\[F = QE\]

Electric Potential:

Electric potential is more commonly known as voltage. The potential at a point a distance r from a charge q is given by: \[V = \dfrac{k q}{r}\] Electric potential, like potential energy, is a scalar, not a vector.

SI unit for electric potential:

1 Volt = 1 Joule/1 Coulomb or \[V = \dfrac{J}{C}\] (The number of volts indicates the number of joules work or energy per coulomb.)

Potential Difference:

Electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy.
In equation form, the electric potential difference is: \[\Delta V = V_B - V_A = \dfrac{W_{B \rightarrow A}}{q}\] Conventional current flows around a circuit from the positive (+) side of the cell to the negative (-). However the electrons are flowing around the circuit in the opposite direction from the negative (-) side of the cell to the positive (+).

Energy Stored in a Capacitor

\[E = \dfrac{1}{2}CV^2 = \dfrac{QV}{2} = \dfrac{Q^2}{2C}\] where Q is the charge and V the voltage on a capacitor C.

Solved Example: 18-2-01

Energy stored in a condenser of capacity 10 $\mu F$, charged to 6 kV is used to lift a mass of 10g . The height to which the body can be raised is:

A. 180 m

B. 18 m

C. 1.8 m

D. 1800 m

Energy stored in a condenser $E = \dfrac{1}{2} CV^2 = \dfrac{1}{2} \times 10 \times 10^{-6} \times 6000^2 = 180\ J$
This will be equal to the potential energy. $180 = mgh, 180 = 0.010 \times 9.81 \times h, h = 1834\ m$

Correct Answer: D

Solved Example: 18-2-02

Two capacitors of capacities 1 $\mu$F and 4 $\mu$F are connected in series with battery of 200 V. The voltage across them are in the ratio of:

A. $\dfrac{1}{2}$

B. $\dfrac{4}{1}$

C. $\dfrac{2}{1}$

D. $\dfrac{1}{4}$

Capacitors in Series will have effective capacitance as: \[\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2}= \dfrac{1}{1} + \dfrac{1}{4}= \dfrac{5}{4}\] \[C = \dfrac{4}{5} = 0.8\ \mu F\] Charge on the effective capacitor is given by, $Q = CV = 0.8 \times 10^{-6} \times 200 = 160 \times 10^{-6}\ \mathrm{Coulombs}$
Capacitors in series will have same charge as that of equivalent. $Q = Q_1 = Q_2 = 160 \times 10^{-6}\ \mathrm{Coulombs}$
\[V_1 = \dfrac{Q_1}{C_1} = \dfrac{160 \times 10^{-6}}{1 \times 10^{-6}} = 160\ V\] \[V_2 = \dfrac{Q_2}{C_2} = \dfrac{160 \times 10^{-6}}{4 \times 10^{-6}} = 40\ V\] \[\dfrac{V_1}{V_2} = \dfrac{160}{40} = \dfrac{4}{1}\]

Correct Answer: B

Solved Example: 18-2-03

How do you arrange four equal capacitors of 4 $\mu F$ to get effective capacity of 3 $\mu F$ ?

A. Three in series, one in parallel

B. Two in parallel, two in series

C. Three in parallel, one in series

D. All four in series

If three capacitors are in series, then effective capacitance = 12 $\mu$ F. This is in series with 4 $\mu$ F. So effective capacitance $\dfrac {1}{C_{\mathrm{eff}}}=\dfrac {1}{12}+\dfrac {1}{4}=\dfrac {1+3}{12}=\dfrac {4}{12}, \quad C_{\mathrm{eff}}=\dfrac {12}{4}=3\mu F$

Correct Answer: C

Solved Example: 18-2-04

In an experiment, a charged drop of oil of mass 1.8 $\times 10^{-14}$ kg is stationary between the plates of a parallel plate condenser separated by a distance of 9 mm and charged to a potential difference of 2000 V. The charge on the drop is:

A. 8 $\times 10^{-19}$ C

B. 6 $\times 10^{-19}$ C

C. 5 $\times 10^{-19}$ C

D. 4 $\times 10^{-19}$ C

Vertical Force = qE
But \[E = \dfrac{V}{d} = \dfrac{2 \times 10^3}{9 \times 10^{-3}} = 0.22 \times 10^6 V/m\] Also, qE = mg \[q = \dfrac{mg}{E} =\dfrac{1.8 \times 10^{-14} \times 9.8}{0.22 \times 10^6} =8.02 \times 10^{-19} C\]

Correct Answer: A

Solved Example: 18-2-05

Point charge q moves from point P to point S along the path PQRS (figure shown) in a uniform electric field E pointing coparallel to the positive direction of the X-axis. The coordinates of the points P,Q,R and S are (a,b), (2a,0), (a,-b) and (0,0) respectively. The work done by the field in the above:

18-2-05

A. qEa

B. qE$\sqrt{(2a)^2+b^2}$

C. qEa$\sqrt{2}$

D. -qEa

18-2-05

As electric field is a conservative field Hence the work done does not depend on path. Instead of following red path, we can follow blue path as shown in the figure, which is much simpler.

\begin{align*} W_{PQRS} &= W_{POS} = W_{PO} + W_{OS}\\ &=Fb \cos 90^\circ +Fa \cos180^\circ\\ &=0 + qEa(1)\\ &=qEa \end{align*}

Correct Answer: A

Solved Example: 18-2-06

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in:

A. Reduction of charge on the plates and increase of potential difference across the plates.

B. Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates.

C. Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates.

D. None of the above.

Battery in disconnected so Q will be constant as C $\propto$ K. So with introduction of dielectric slab capacitance will increase using Q = CV, V will decrease and using U= $\dfrac{Q^2}{2C}$, energy will decrease.

Correct Answer: C

Learning Objectives:

Define electric current.
Calculate electric current from charge and time.

Electric current is defined as the rate at which charge flows through a surface (such as cross section of a wire).

\[di(t) = \dfrac{dq(t)}{dt}\]

Or in stedy state format, \[I = \dfrac{Q}{t}\] I = electric current rate at which charge flows in a certain pathway
Q = amount of charge flowing past a certain point
t = time
SI unit for electric current:
1 Ampere = 1 Coulomb/1 second

Solved Example: 18-3-01

Rate of flow of charge through cross-sectional area is known as: (MP Aug 2017- Shift I)

A. Current

B. Voltage

C. Electric potential

D. Charge velocity

Correct Answer: A

Solved Example: 18-3-02

Current flowing through a conductor, when 2 $\times$ 10$^7$ electrons pass in 1 $\mu$sec will be:

A. 1.6 $\times$ 10$^{-6}$

B. 3.2 $\times$ 10$^{-6}$

C. 4.2 $\times$ 10$^{-6}$

D. 6 $\times$ 10$^{-6}$

Each electron has a charge of 1.6 $\times$ 10$^{-19}$ C.
\begin{align*} \mathrm{current} = \dfrac{\mathrm{charge}}{\mathrm{time}} &= \dfrac{1.6 \times 10^{-19} \times 2 \times 10^7}{1 \times 10^{-6}}\\ &= 3.2 \times 10^{-6}\ \mathrm{A} \end{align*}

Correct Answer: B

Learning Objectives:

Define the concept of power and understand the relationship between current, voltage, and power.
Calculate power in electric circuits (both electrical energy transfer and energy lost through resistance)
Understand heating effect of electric current.

Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit.
The SI unit of power is the Watt, one Joule per Second.

\[P = V I = R I^2 = \dfrac{V^2}{R} = \dfrac{VQ}{t}\]

Solved Example: 18-5-01

If we double both current and voltage keeping resistance constant then power is:

A. Unchanged

B. Halves

C. Doubles

D. Quadruples

\[P = V_{new} \cdot I_{new} = 2V_{old} \cdot 2I_{old} = 4P_{old}\]

Correct Answer: D

Solved Example: 18-5-02

If a light bulb is switched on for 40 s and it consumes 2400 J of electrical energy then its power is:

A. 60 W

B. 70 W

C. 80 W

D. 90 W

$P = \dfrac{W}{t} = \dfrac{2400}{40} = 60\ W$

Correct Answer: A

Solved Example: 18-5-03

Amount of energy supplied by current in unit time is called:

A. Electric energy

B. Electric power

C. Resistance

D. Friction

Rate of energy is power. If the power is given by displacement of body, it is mechanical power. If it is given by movement of electrons (in the form of current), it is electrical power.

Correct Answer: B

Solved Example: 18-5-04

If resistance of an electric bulb is 500 $\Omega$ and voltage across its ends is 250 V then power consumed by it is:

A. 130 W

B. 125 W

C. 120 W

D. 200 W

$P = V \cdot I = V \cdot \left(\dfrac{V}{R}\right) = \dfrac{V^2}{R} = \dfrac{250^2}{500} = 125\ W$

Correct Answer: B

Electrical Fundamentals