First slide

Capacitance

Question

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field due to a point charge Q(having a charge equal to the sum of the charges on the 4 $μ$ F and 9 $μ$ F capacitors), at a point distance 30m from it, would equal :

Easy

A

360 N/C
B

420 N/C
C

480 N/C
D

240 N/C

Solution

$\begin{aligned} Q_{1} = 24 μC \\ Q_{2} = 18 μC \end{aligned} Q_{3} = 42 μC = Q_{1} + Q_{2} E = 10^{7} \times 42 \times 10^{- 6} (E = \frac{kq}{r^{2}}) E = 420 N / C$

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