As a charged particle travels from a point at a potential of 100 V to a point at 200 V , its kinetic energy decreases by 10 J . Determine the charge on the particle.

Correct answer: B. 0.1 C Using the principle of conservation of energy: Δ K + Δ U = 0 Given that the kinetic energy decreases by 10 J , we have Δ K = -10 J . Therefore, the change in potential energy is: Δ U = -Δ K = 10 J The change in potential energy is also given by the relation: Δ U = q(V_f - V_i) Substituting the given values V_i = 100 V and V_f = 200 V : 10 = q(200 - 100) 10 = 100q q = 0.1 C

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Electric Field and Potential

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Field and Potential

As a charged particle travels from a point at a potential of 100 V...

As a charged particle travels from a point at a potential of

100 \text{ V}

to a point at

200 \text{ V}

, its kinetic energy decreases by

10 \text{ J}

. Determine the charge on the particle.

Options

A $1000 \text{ C}$
B $0.1 \text{ C}$
C $10 \text{ C}$
D $-0.1 \text{ C}$

Correct answer

B. $0.1 \text{ C}$

Step-by-step solution

Using the principle of conservation of energy:

\Delta K + \Delta U = 0

Given that the kinetic energy decreases by

10 \text{ J}

, we have

\Delta K = -10 \text{ J}

.

Therefore, the change in potential energy is:

\Delta U = -\Delta K = 10 \text{ J}

The change in potential energy is also given by the relation:

\Delta U = q(V_f - V_i)

Substituting the given values

V_i = 100 \text{ V}

and

V_f = 200 \text{ V}

10 = q(200 - 100)

10 = 100q

q = 0.1 \text{ C}

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