Question

A – 41 nC charge is placed in between two-point charges, 32 nC and 24 nC. The negative

charge moves from point a to b. What is the change in potential energy?

Answer 1

Solution;

Find the potential at point a;

Due to 32nC;

$U_{32,-41}=\frac{kq_1q_2}{r}$

$U=\frac{8.99×10^9×32×-41×10^{-18}}{0.1}=1.179×10^{-4}J$

Due to the 24nC;

$U=\frac{8.99×10^9×-41×24×10^{-18}}{0.05}=1.769×10^{-4}J$

Total potential energy at point a is;

$U_a=(1.179+1.769)×10^{-4}=2.948×10^{-4}J$

After moving to b;

Distance between the 32nC is;

$r=\sqrt{10^2+7^2}=0.122m$

Potential energy becomes;

$U=\frac{8.99×10^9×41×32×10^{-18}}{0.122}=0.9667×10^{-5}J$

Distance between the 25nC becomes;

$r=\sqrt{5^2+7^2}=0.0862$

Potential energy becomes;

$U=\frac{8.99×24×41×10^{-18}}{0.0862}=1.0262×10^{-4}J$

Total potential energy at point b is;

$U_b=(1.0262+0.9667)×10^{-4}=1.9929×10^{-4}J$J

The difference in potential difference is;

$\Delta U=U_a-U_b$

$\Delta U=2.984-1.9924=0.9911×10^{-4}J$