Q174 / 180 · COMEDK 2015 · Physics
Two small charged spheres $A$ and $B$ have charges $10 \mu \mathrm{C}$ and $40 \mu \mathrm{C}$ respectively and are held at separation of $90 \mathrm{~cm}$ from each other. At what distance from $A$, electric field intensity would be zero?
- A$22.5 \mathrm{~cm}$
- B$18 \mathrm{~cm}$
- Correct answerC$30 \mathrm{~cm}$
- D$36 \mathrm{~cm}$
Solution
The given situation is shown below,

Let electric field at point $P$ at a distance $x$ from $A$ is zero. At point $P, E_{A}=E_{B}$ $\Rightarrow \quad \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A}}{x^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{B}}{(r-x)^{2}}$ $\Rightarrow \quad \frac{10 \times 10^{-6}}{x^{2}}=\frac{40 \times 10^{-6}}{(r-x)^{2}}$ $\Rightarrow \quad\left(\frac{1}{x}\right)^{2}=\left(\frac{2}{r-x}\right)^{2}$ $\Rightarrow \quad \frac{1}{x}=\frac{2}{r-x}$ $\rightarrow \quad 2 x-r-x$ $\Rightarrow \quad x=\frac{r}{3}$ $$ =\frac{90}{3}=30 \mathrm{~cm} $$

Let electric field at point $P$ at a distance $x$ from $A$ is zero. At point $P, E_{A}=E_{B}$ $\Rightarrow \quad \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A}}{x^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{B}}{(r-x)^{2}}$ $\Rightarrow \quad \frac{10 \times 10^{-6}}{x^{2}}=\frac{40 \times 10^{-6}}{(r-x)^{2}}$ $\Rightarrow \quad\left(\frac{1}{x}\right)^{2}=\left(\frac{2}{r-x}\right)^{2}$ $\Rightarrow \quad \frac{1}{x}=\frac{2}{r-x}$ $\rightarrow \quad 2 x-r-x$ $\Rightarrow \quad x=\frac{r}{3}$ $$ =\frac{90}{3}=30 \mathrm{~cm} $$
Hence, the correct answer is option: 'C'.