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konstantin123 [22]
3 years ago
11

A 5.0 c charge is 10 m from a small test charge. what is the magnitude of the electric field at the location of the test charge

?
Physics
1 answer:
Inessa [10]3 years ago
7 0

Answer:

4.50*10^8\frac{N}{C}

Explanation:

The electric field is generated by a charge and represents the force exerted on a test charge, that is, the electric force per unit of charge. Therefore the equation for the electric field can be obtained from Coulomb's law.

E=\frac{F}{q}\\E=\frac{kq}{r^2}\\E=\frac{5C*8.99*10^9\frac{Nm^2}{C^2}}{(10 m)^2}=4.50*10^8\frac{N}{C}

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\rho = 12580.7 kg/m^3

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As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

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