Question 450: Electrostats & Capacitance

Question

Two identical metal plates are given charges $q_1$ and $q_2\left(q_2 < q_1\right)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance ' C ', the potential difference ' $V$ ' between the plates is

Accepted Answer

The effective charge on the capacitor is:
$$ Q=\frac{q_1-q_2}{2} $$
From formula for capacitance,
$$ \begin{aligned} & V=\frac{q}{C}=\frac{q_1-q_2}{2 C} \ 	herefore & V=\frac{q_1-q_2}{2 C} \end{aligned} $$

Answer

$\frac{q_1-q_2}{c}$

Answer

$\frac{q_1+q_2}{C}$

Answer

$\frac{q_1-q_2}{2 C}$

Answer

$\frac{q_1+q_2}{2 C}$

Electrostats & Capacitance

Two identical metal plates are given charges q_1 and q_2\left(q_2 < q_1\right) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance ' C ', the potential difference ' V ' between the plates is

Choose the correct answer: