Question

Given the potential $V=\frac{10}{r^2}\sin \theta \cos \varphi$. Calculate the work done in moving a $10\mu C$ charge from point $A(1,{30}^o,{120}^o)$ to $B(4,{90}^o,{60}^o)$ :

• A. $28.125J$
• B. $28.125mJ$
• C. $28.125\mu J$
• D. $28.125pJ$

Answer 1

The correct option is C $28.125\mu J$

We know that, in the presence of conservative forces, the net work done is equal to the change in potential energy of the system.

The potential energy of the charge at $A$ is ${\text{PE}}_i=q{V}_A$

The potential energy of the charge at $B$ is ${\text{PE}}_f=q{V}_B$

Thus, work done is $W={\text{PE}}_f-{\text{PE}}_i=q(V_B-V_A)$

Given that $q=10\mu \text{C}$

$V_A=\frac{10}{1^2}\sin \left({30}^{\circ }\right)\cos \left({120}^{\circ }\right)=10\times \frac{1}{2}\times \frac{-1}{2}=-2.5\text{J/C}$

$V_B=\frac{10}{4^2}\sin \left({90}^{\circ }\right)\cos \left({60}^{\circ }\right)=\frac{10}{16}\times 1\times \frac{1}{2}=0.3125\text{J/C}$

Thus, $W=q(V_B-V_A)=10\mu \text{C}\times (0.3125+2.5)\text{J/C}=28.125\mu \text{J}$