Electrostatics Ques 174
- Four point charges, each of $+q$, are rigidly fixed at the four corners of a square planar soap film of side $a$. The surface tension of the soap film is $\gamma$. The system of charges and planar film are in equilibrium, and $a=k{\frac{q^{2}}{\mathrm{Y}}}^{1 / N}$, where $k$ is a constant. Then $N$ is
(2011)
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Answer:
Correct Answer: 174.3
Solution:
Formula:
Coulomb Force Between Two Point Charges:
$F_{1}=$ Net electrostatic force on any one charge due to rest of three charges
$$ =\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{a^{2}} \quad \sqrt{2}+\frac{1}{2} $$
$F_{2}=$ Surface tension force $=\gamma a$
If we consider the equilibrium of line $BC$, then
$$ \begin{gathered} 2 F_{1} \cos 45^{\circ}=F_{2} \\ \text { or } \sqrt{2} F_{1}=F_{2} \\ \text { or } \quad \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{a^{2}} + \frac{1}{\sqrt{2}}=\gamma a \\ \therefore \quad a^{3}=\frac{1}{4 \pi \varepsilon_{0}} \cdot 2+\frac{1}{\sqrt{2}} \frac{q^{2}}{\mathrm{Y}} \ \text { or } \quad a=\frac{1}{4 \pi \varepsilon_{0}} \cdot 2+\frac{1}{\sqrt{2}} \quad \frac{q^{2}}{\mathrm{Y}}=k \frac{q^{2}}{\mathrm{Y}} \end{gathered} $$
where, $k=\frac{1}{4 \pi \varepsilon_{0}} \quad 2+\frac{1}{\sqrt{2}}$
Therefore, $N=3$
Answer is 3 .
BETA
