Q19 / 142 · NTA Abhyas (JEE Main) 2020 · Physics · Electrostatics

In the given figure the work done by the battery after the switch $S$ is closed is

A
$100 μ J$
B
$- 100 μ J$
Correct answer
C
$80 μ J$
D
$- 80 μ J$

Solution

Before the switch S is closed

Q_{i} = C_{eq} V

= 5 \times 4 = 20 μ C

When the switch S is closed

Q_{f} = CV

= 10 \times 4 = 40 μ C

W (cell) = Δ QE = (Q_{f} - Q_{i}) \times E

= 20 \times 4 = 80 μ J

Hence, the correct answer is option: 'C'.

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