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insens350 [35]
3 years ago
15

As a particle moves 10.0 m along an electric field of strength 75 N/C, its electrical potential energy decreases by 4.8 × 10-16

J. What is the particle's charge?
Physics
1 answer:
Aleks04 [339]3 years ago
6 0

Answer:

6.4×10⁻¹⁹ C

Explanation:

From the question,

V = E×d.............. Equation 1

Where V = electric potential, E = Electric Field, d = distance moved by the particle.

Given: E = 75 N/C, d = 10 m.

Substitute into equation 1

V = 75×10 = 750 V.

Also Using,

W = qV.................. Equation 2

W = Work done by the particle when it moves, q = Particles charge

make q the subject of the equation,

q = W/V............ Equation 3

Given: W = 4.8×10⁻¹⁶ J, V = 750 V

Substitute into equation 3

q = 4.8×10⁻¹⁶/750

q = 6.4×10⁻¹⁹ C

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The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on
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Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 11 cm that is placed in a spatially uniform m
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Answer:

Explanation:

Given that,

Number of turn N = 40

Diameter of the coil d= 11cm = 0.11m

Then, radius = d/2 = 0.11/2 =0.055m

r = 0.055m

Then, the area is given as

A =πr²

A = π × 0.055²

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E.M.F is given as

ε = —N • dΦ/dt

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Then, ε = —N • dΦ/dt

ε = —N • dBA/dt

ε = —NBA/t

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3 years ago
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