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NemiM [27]
3 years ago
5

The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol

t-cubic meter. As a multiple of A, what is the magnitude of the electric potential difference between r1 = 1.71 m and r2 = 2.89 m?
Physics
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

\Delta V = 0.053 A

Explanation:

Electric field in a given region is given by equation

E = \frac{A}{r^4}

as we know the relation between electric field and potential difference is given as

\Delta V = -\int E. dr

so here we have

\Delta V = - \int (\frac{A}{r^4}) .dr

\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}

here we know that

r_1 = 1.71 m  and r_2 = 2.89 m

so we will have

\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})

so we will have

\Delta V = 0.053 A

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