### 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.

### 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:

Solution:
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. $180 = mgh, 180 = 0.010 \times 9.81 \times h, h = 1834\ m$

### 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:

Solution:
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. $Q = Q_1 = Q_2 = 160 \times 10^{-6}\ \mathrm{Coulombs}$
$V_1 = \dfrac{Q_1}{C_1} = \dfrac{160 \times 10^{-6}}{1 \times 10^{-6}} = 160\ V$ $V_2 = \dfrac{Q_2}{C_2} = \dfrac{160 \times 10^{-6}}{4 \times 10^{-6}} = 40\ V$ $\dfrac{V_1}{V_2} = \dfrac{160}{40} = \dfrac{4}{1}$

### Solved Example:

#### 18-2-03

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

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

### 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:

Solution:
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 $q = \dfrac{mg}{E} =\dfrac{1.8 \times 10^{-14} \times 9.8}{0.22 \times 10^6} =8.02 \times 10^{-19} C$

### 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: Solution: 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*}

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.