Question

Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C. where, d= 11 cm Q= 12.5 C

Answer 1

Answer:

The electric field is 9.3 x 10^12 N/C and the direction is away from the charge.

Explanation:

charge, Q = 12.5 C

distance, d = 11 cm = 0.11 m

Let the electric field is E.

$E =\frac{K Q}{d^2}\\\\E = \frac{9\times 10^9\times 12.5}{0.11\times 0.11}\\\\E = 9.3\times 10^{12} N/C$

The direction of electric filed is away from the charge.