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 4µF and 9µF capacitors), at a point distance 30 m from it, would equal:
Text Solution
Verified by ExpertsThe correct answer is:
B
Q 1 = 24µc
Q 2 = 18 µc
Q = 42µc
E = 10 7 × 42 × 10 – 6
E = 420 N/C
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