Question

Two point charges $ + 10^{-7} C $ and $ - 10^{-7} C $ are placed at $ A $ and $ B, 20 \,cm $ apart as shown in the figure. Calculate the electric field at $ C, \,20\, cm $ apart form both $ A $ and $ B $ .

• A. $ 1.5 \times 10^{-5} N/C $
• B. $ 2.2 \times 10^{4} N/C $
• C. $ 3.5 \times 10^{6} N/C $
• D. $ 3.0 \times 10^{5} N/C $

Answer 1

Answer: (B)

The magnitude of each electric field vector at point $C$ due to charge $q_1$ and $q_2$
$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2}$
$E = \frac{9 \times 10^9 \times 10^{-7}}{(20 \times 10^{-2})^2}$
$ = 2.2 \times 10^4 \,N/C$