Electrostatics Ques 27
- Three charges $+Q, q,+Q$ are placed respectively at distance $0, \frac{d}{2}$ and $d$ from the origin on the $X$-axis. If the net force experienced by $+Q$ placed at $x=0$ is zero, then value of $q$ is
(a) $\frac{+Q}{2}$
(b) $\frac{+Q}{4}$
(c) $\frac{-Q}{2}$
(d) $\frac{-Q}{4}$
(Main 2019, 9 Jan Shift I)
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Answer:
Correct Answer: 27.(d)
Solution:
Formula:
Coulomb Force Between Two Point Charges:
- The given condition is shown in the figure given below,
Then, according to the Coulomb’s law, the electrostatic force between two charges $q _1$ and $q _2$ such that the distance between them is $(r)$ given as,
$$ F=\frac{1 \cdot q _1 q _2}{4 \pi \varepsilon _0 \cdot r^{2}} $$
$\therefore$ Net force on charge ’ $Q$ ’ placed at origin i.e. at $x=0$ in accordance with the principle of superposition can be given as
$$ F _{net}=\frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q \times q}{\frac{d}{2}}+\frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q \times Q}{(d)^{2}} $$
Since, it has been given that, $F _{\text {net }}=0$.
$$ \begin{aligned} & \Rightarrow \quad \frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q \times q}{\frac{d}{2}}+\frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q \times Q}{(d)^{2}}=0 \\ & \Rightarrow \quad \frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q \times q}{\frac{d^{2}}{2}}=-\frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q \times Q}{(d)^{2}} \text { or } q=-\frac{Q}{4} \end{aligned} $$
BETA
