Question

^{charge lie in a straight line with -3nC 2cm from point p and 2nc is 1 cm from p which is 2cm from point A.}

^calculate

a) potential between p and q

b) potential difference between p and q

c)work done to take 3nC from q to p

d)gain energy of electrons moving freely from q to p in joules an eV

Answer 1

a)

$$V(x)=-\frac{(9\cdot10^{9})(3\cdot10^{-9})}{0.02+x}+\frac{(9\cdot10^{9})(2\cdot10^{-9})}{0.01+x}\\ V(x)=-\frac{27}{0.02+x}+\frac{18}{0.01+x}$$

b)

$$V(0)-V(0.02)\\=-\frac{27}{0.02}+\frac{18}{0.01}+\frac{27}{0.02+0.02}-\frac{18}{0.01+0.02}=525 V$$

c)

$$W=525(3\cdot10^{-9})=1.575\cdot10^{-6}\ J$$

d)

$$W_e=525(1.6\cdot10^{-19})=8.4\cdot10^{-17}\ J=525\ eV$$