Answer to Question #201282 in Field Theory for Kenneth

Question

Given that a positive charge Q1=+3c and negative charge Q2=2c, are separated by a distance of 8cm. Take permittivity constant k=9.0×10^9NM^2/c^2.

(a). find The electric field at point p 1cm from Q2

(b) what is the electric potential at point P ,5 cm from Q1.

Answer 1

(a) The electric field at point P 1cm from Q₂ can be found as follows:

$$E=E_1-E_2=k(\dfrac{Q_1}{(x+d)^2}-\dfrac{Q_2}{x^2}),$$ $$E=9\cdot10^9\ \dfrac{N\cdot m^2}{C^2}\cdot(\dfrac{3\cdot10^{-6}\ C}{(0.01\ m+0.08\ m)^2}-\dfrac{2\cdot10^{-6}\ C}{(0.01\ m)^2}),$$ $$E=-1.76\cdot10^{8}\ \dfrac{N}{C}.$$

The sign minus means that the net electric field at point P directed leftward (towards the charge Q₁).

(b) The electric potential at point P 5cm from Q₁ can be found as follows:

$$V=V_1+V_2,$$ $$V=\dfrac{kQ_1}{x}+\dfrac{kQ_2}{d+x}=k(\dfrac{Q_1}{x}+\dfrac{Q_2}{d+x}),$$ $$V=9\cdot10^9\ \dfrac{N\cdot m^2}{C^2}\cdot(\dfrac{3\cdot10^{-6}\ C}{0.05\ m}+\dfrac{-2\cdot10^{-6}\ C}{0.05\ m+0.08\ m}),$$ $$V=4\cdot10^5\ V.$$