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Bond [772]
4 years ago
8

The electric field 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's char

ge?
Physics
1 answer:
Sonbull [250]4 years ago
4 0

Answer:

2.1×10¹⁸ C

Explanation:

Using,

E = kq/r²...................... Equation 1

Where E = Electric field, q = charge, r = distance, k = coulombs constant.

make q the subject of the equation

q = Er²/k.................. Equation 2

Given: E = 180000 N/C, r = 2.8 cm = 0.028 m

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

q = 180000(9×10⁹)/0.028²

q = 2.1×10¹⁸ C

Hence the object charge is 2.1×10¹⁸ C

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