Electric Field and Electric Potential

  • 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 Examples

Solved Example:


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:

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:


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:

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:


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

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:


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:

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:


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:




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:


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

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