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aev [14]
3 years ago
13

Points P and Q are separated by a distance of 0.10 m in a uniform electric field. The potential difference between points P and

Q is 55 V. What is the magnitude of the electric field?
Physics
1 answer:
Vadim26 [7]3 years ago
7 0

Answer:

E = 550 V/m = 550 N/C

Explanation:

The formula for calculating the potential difference between two points in an electric field is given as follows:

ΔV = E.Δr

where,

ΔV = Potential Difference between points P and Q = 55 V

Δr = Distance between P and Q = 0.1 m

E = Magnitude of Electric Field = ?

Therefore, using these values in the equation, we get:

55 V = E(0.1 m)

E = 55 V/0.1 m

<u>E = 550 V/m = 550 N/C</u>

Because, volt/meter is equivalent to Newton/Coulomb.

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