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Zepler [3.9K]
4 years ago
11

A charge of 9.0 × 10^–5 C is placed in an electric field with a strength of 2.0 × 10^4 N/C. What is the electrical force acting

on the charge?
a) What is given?
b) What are you asked to find?
c) What equation should you use?
d) What is the solution?

please help​

Physics
1 answer:
Tresset [83]4 years ago
7 0

Answer: the force = 1.8N

check the pic for explanation

Explanation:

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A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational
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The universal law of gravitation is defined as:

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Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em>m = 5.0 Kg<em>:</em>

m1 will be equal to the mass of the Earth m1 = 5.972×10^{24} Kg and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth r = 6371000m.

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}

F = 49.06N

Newton's second law can be used to know the acceleration.

F = ma

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a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}

a =9.8m/s^{2}

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F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}

F = 29.44N

a =\frac{F}{m}

a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}

a =9.8m/s^{2}

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