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Physics Question on Electric Field

The electric field lines on the left have twice the separation on those on the right as shown in figure. If the magnitude of the field at AA is 40 Vm−140\,Vm^{-1}, what is the force on 20 μ C20\,\mu \,C charge kept at BB?

A

4×10−4Vm−14 \times 10^{-4} Vm^{-1}

B

8×10−4Vm−18 \times 10^{-4} Vm^{-1}

C

16×10−4Vm−116 \times 10^{-4} Vm^{-1}

D

1×10−4Vm−11 \times 10^{-4} Vm^{-1}

Answer

4×10−4Vm−14 \times 10^{-4} Vm^{-1}

Explanation

Solution

We know that F=qE=20×20×10−6F = qE =20 \times 20 \times 10^{-6} =4×10−4V/m−1=4 \times 10^{-4} V / m^{-1}