At a distance of 40 cm from a point charge, the electric field has a magnitude of 5.0 N C^ -1 . Determine the magnitude of the charge.

Correct answer: D. 8.9 × 10^ -11 C The electric field due to a point charge q at a distance r is given by E = (1)/(4 π _0) (q)/(r^2) . Given E = 5.0 N C^ -1 and r = 40 cm = 0.4 m . Substituting the values: 5.0 = (9 × 10^9 × q)/((0.4)^2) q = (5.0 × 0.16)/(9 × 10^9) q = (0.8)/(9 × 10^9) q = 8.88 × 10^ -11 C 8.9 × 10^ -11 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

At a distance of 40 cm from a point charge, the electric field has...

At a distance of

40 \text{ cm}

from a point charge, the electric field has a magnitude of

5.0 \text{ N C}^{-1}

. Determine the magnitude of the charge.

Options

A $2.2 \times 10^{-10} \text{ C}$
B $8.9 \times 10^{-7} \text{ C}$
C $8.9 \times 10^{-9} \text{ C}$
D $8.9 \times 10^{-11} \text{ C}$

Correct answer

D. $8.9 \times 10^{-11} \text{ C}$

Step-by-step solution

The electric field due to a point charge

q

at a distance

r

is given by

E = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q}{r^2}

.

Given

E = 5.0 \text{ N C}^{-1}

and

r = 40 \text{ cm} = 0.4 \text{ m}

.

Substituting the values:

5.0 = \dfrac{9 \times 10^9 \times q}{(0.4)^2}

q = \dfrac{5.0 \times 0.16}{9 \times 10^9}

q = \dfrac{0.8}{9 \times 10^9}

q = 8.88 \times 10^{-11} \text{ C} \approx 8.9 \times 10^{-11} \text{ C}

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