If the accelerating plates are 1.70 cm apart, and have a potential difference of

Question

If the accelerating plates are 1.70 cm apart, and have a potential difference of 2.10×104 V , what is the magnitude of the uniform electric field between them?

The Electric Potential of the Earth The Earth has a vertical electric field with a magnitude of approximately 100 V/m near its surface.What is the magnitude of the potential difference between a point on the ground and a point on the same level as the top of the Washington Monument (555 ft high)?

How far must the point charges q1 = 7.30 C and q2 = -25.0 C be separated for the electric potential energy of the system to be -150 J ?

Explanation / Answer

1)
d = 1.7 cm = 0.017 m
V = 2.10*10^4 V
E = V/d
   = (2.10*10^4)/0.017
   = 1.24*10^6 V/m
2)
1ft = 0.3048 m
d = 555 ft = 555*0.3048 m = 169.164 m
E = 100 V/m
V = E*d
    = 100 * 169.164
   = 1.69*10^4 V

I am allowed to answer only 1 question at a time but I have answered 2

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