Question

Two point charges $q_1$ and $q_2$ of magnitude $+{10}^{-8}$ and $-{10}^{-8}C$, respectively, are placed 0.1m apart. Calculate the magnitude of electric field at point A, shown in figure.

• A. $9\times {10}^{3}N/C$
• B. $5\times {10}^{3}N/C$
• C. $3\times {10}^{3}N/C$
• D. $11\times {10}^{3}N/C$

Answer 1

The correct option is A $9\times {10}^{3}N/C$

We evaluate the electric field at the point in x direction and y direction. Clearly, electric field in x direction is zero because of symmetry. In y direction, apply the superposition principle to get,
$E_A=2\left(\frac{k{q}_1}{(0.1{)}^2}\right)cos{60}^o=9\times {10}^{3}N/C$