Question

Kinetic energy of a charged particle decreases by 10 J as it moves from a point at potential 100 v to 200 v. The charge on the particle is:
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Answer 1

Given:

Decrease in kinetic energy (∆K) = 10 J

Potential of points between which charge particle moves:

$\rm V_1 = 100 \ V \\ \rm V_2 = 200 \ V$

To Find:

Charge on the particle (q)

Answer:

From conservation of energy:

Increase in potential energy = Decrease in kinetic energy

∆U = ∆K = 10 J

Electric potential difference is the work done by external force in moving a unit positive charge from one point to other.

$\boxed{ \bf{\Delta V = \dfrac{\Delta U}{q}}}$

$\rm \implies q = \dfrac{\Delta U}{\Delta V} \\ \\ \rm \implies q = \dfrac{\Delta U}{V_2 - V_1} \\ \\ \rm \implies q = \dfrac{10}{200 - 100} \\ \\ \rm \implies q = \frac{ \cancel{10}}{10 \cancel{0}} \\ \\ \rm \implies q = \frac{1}{10} \: C\\ \\ \rm \implies q = 0.1\: C$

$\therefore$ $\boxed{\mathfrak{Charge \ on \ the \ particle \ (q) = 0.1 \ C}}$