Question #201282

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.



1
Expert's answer
2021-06-01T10:50:38-0400

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


E=E1E2=k(Q1(x+d)2Q2x2),E=E_1-E_2=k(\dfrac{Q_1}{(x+d)^2}-\dfrac{Q_2}{x^2}),E=9109 Nm2C2(3106 C(0.01 m+0.08 m)22106 C(0.01 m)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=1.76108 NC.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=V1+V2,V=V_1+V_2,V=kQ1x+kQ2d+x=k(Q1x+Q2d+x),V=\dfrac{kQ_1}{x}+\dfrac{kQ_2}{d+x}=k(\dfrac{Q_1}{x}+\dfrac{Q_2}{d+x}),V=9109 Nm2C2(3106 C0.05 m+2106 C0.05 m+0.08 m),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=4105 V.V=4\cdot10^5\ V.

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