Electrostatic Potential and Work Done Trick

 

❓ Question

Find the work done in bringing a charge

$$q=3\,\text{nC}$$

from infinity to point $A$.

Given:

• $QB=1\,\text{nC}$
  
• $QC=3\,\text{nC}$
  
• $AB=AC=BC=3\,\text{cm}$
  

(All three sides are $3\ \text{cm}$.)

---

✍️ Solution

Step 1: Potential at point $A$

Potential due to a point charge:

$$V=\frac{kq}{r}$$

Since both charges are at the same distance from $A$,

$$r=3\text{ cm}=3\times10^{-2}\text{ m}$$

Hence,

$$V_A=\frac{kq_B}{r}+\frac{kq_C}{r}$$
$$=\frac{k(q_B+q_C)}{r}$$

Substitute values:

$$k=9\times10^9, \quad q_B=1\times10^{-9}\text{ C}, \quad q_C=3\times10^{-9}\text{ C}$$
$$V_A = \frac{9\times10^9(4\times10^{-9})}{3\times10^{-2}}$$
$$= \frac{36}{0.03} = 1200\ \text{V}$$

---

Step 2: Work done in bringing the charge

Work done by an external agent:

$$W=qV$$

Given,

$$q=3\times10^{-9}\text{ C}$$

Therefore,

$$W = 3\times10^{-9}\times1200$$
$$= 3600\times10^{-9} = 3.6\times10^{-6}\ \text{J}$$

---

✅ Final Answer

$$\boxed{3.6\times10^{-6}\ \text{J}}$$

---

Common Mistake

The handwritten solution writes the final answer as $36\times10^{-7}\,\text{J}$, which is equivalent to

$$36\times10^{-7}=3.6\times10^{-6}\ \text{J}.$$

So the numerical answer is correct.

📚 Related Topics

• GP Product and Sum Relation Trick
• Integral Calculus Function Transformation Trick
• Find fogx Using Functional Equation Trick​