Question

In figure, plate $A$ has $100 μ C$ charge, while plate $B$ has $60 μ C$ charge.

When both switches are open, then

$q_{1} =$ ______

$q_{2} =$ ______

$q_{3} =$ ______

$q_{4} =$ ______

Accepted Answer

Given,

Charge on plate $A$ is $100 μC$ ,

That is, $q_{1} + q_{2} = 100 μC . . . (i)$

Charge on plate $B$ is $60 μC$ ,

That is, $q_{3} + q_{4} = 60 μC . . . (ii)$

By induction, the charges on the facing surface of both the plates will be equal and opposite,

i.e., $q_{2} = - q_{3} . . . (iii)$

From the equations $(i)$ , $(ii)$ and $(iii)$ ,

$q_{1} = 100 μC - q_{2}$ ,

$q_{3} = - q_{2}$ and

$q_{4} = 60 μC - q_{3} = 60 μC + q_{2}$ .

Now, consider a point $P$ inside the conducting plate $B$ .

The Electric field inside the conductor is zero. So, the field at point $P$ , $E_{P} = 0$ .

The electric field at point $P$ due to charges at each surface of the conducting plates $A$ and $B$ is

$E_{P} = \frac{q_{1}}{2 A ε_{0}} + \frac{q_{2}}{2 A ε_{0}} + \frac{q_{3}}{2 A ε_{0}} - \frac{q_{4}}{2 A ε_{0}}$

$E_{P} = \frac{(100 μC - q_{2})}{2 A ε_{0}} + \frac{q_{2}}{2 A ε_{0}} + \frac{(- q_{2})}{2 A ε_{0}} - \frac{(60 μC + q_{2})}{2 A ε_{0}}$ ,

(as $E_{P} = 0$ ),

$\Rightarrow 0 = \frac{100 μC}{2 A ε_{0}} - \frac{q_{2}}{2 A ε_{0}} + \frac{q_{2}}{2 A ε_{0}} - \frac{q_{2}}{2 A ε_{0}} - \frac{60 μC}{2 A ε_{0}} - \frac{q_{2}}{2 A ε_{0}}$

$\Rightarrow 0 = \frac{40 μC}{2 A ε_{0}} - \frac{2 q_{2}}{2 A ε_{0}}$

$\Rightarrow 2 q_{2} = 40 μC$

$\Rightarrow q_{2} = 20 μC$ .

Therefore,

$q_{1} = 100 μC - q_{2} = 100 μC - 20 μC = 80 μC$ .

$q_{3} = - q_{2} = - 20 μ C$ and

$q_{4} = 60 μC + q_{2} = 60 μC + 20 μC = 80 μC$ .

B M Sharma Solutions for Chapter: Capacitor and Capacitance, Exercise 5: Concept Application Exercise

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Capacitor and Capacitance, Exercise 5: Concept Application Exercise

Questions from B M Sharma Solutions for Chapter: Capacitor and Capacitance, Exercise 5: Concept Application Exercise with Hints & Solutions

Learn and Prepare for any exam you want !