Question 882: Electrostats & Capacitance

Question

'$$\mathrm{F}$$' is the force between the two identical charged particles placed at a distance '$$\mathrm{Y}$$' from each other. If the distance between the charges is reduced to half the previous distance then force between them becomes

Accepted Answer

$$\begin{aligned} & \mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{\mathrm{r}^2} \ & 	herefore \quad \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}ight)^2=(2)^2=4 \ & 	herefore \quad \mathrm{F}_2=4 \mathrm{~F}_1\end{aligned}$$

Answer

$$\frac{F}{4}$$

Answer

$$4 \mathrm{~F}$$

Answer

$$2 \mathrm{~F}$$

Answer

$$\frac{\mathrm{F}}{2}$$

Electrostats & Capacitance

'$\mathrm{F}' is the force between the two identical charged particles placed at a distance '\mathrm{Y}$' from each other. If the distance between the charges is reduced to half the previous distance then force between them becomes

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