First slide

Electrostatic potential

Question

A charged particle of mass 4 gm and having a charge +1 $μ$ C is fired from a point with speed 10 m/s where electric potential is 4x10⁵ V in the direction where electric potential is decreasing, find the potential at a point (in x 10⁵ volts) where its speed becomes $10 \sqrt{2} m / s$ m/s. Express your answer in multiple of 10⁵.

Moderate

Solution

Work done by electric field = gain in KE of the particle.
$q (V_{2} - V_{1}) = \frac{1}{2} m ({v^{2}}_{2} - {v^{2}}_{1})$
$\begin{array}{l} \Rightarrow 1 \times 10^{- 6} [4 \times 10^{5} - V] = \frac{1}{2} \times 4 \times 10^{- 3} \times [{(10 \sqrt{2})}^{2} - 10^{2}] \\ \Rightarrow V = 2 \times 10^{5} V \end{array}$

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