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 µF and 9 µF capacitors), at a point distant 30 m from it, would equal:
-
240N/C
-
360N/C
-
420N/C
-
480N/C
Solution
C.
420N/C
Resultant circuit,
As, charge on 3μF = 3μF x 8V = 24μC
Charge on 3μF = 3μF x 2V = 18 μC
charge on 4μF +Charge on 9μF
= (24 + 18)μC = 42μC
therefore,
Electric field at a point distant 30 m
