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Ket [755]
3 years ago
15

The potential difference between two equipotential lines 5.0 mm apart has a value of 1.2 V, calculate the electric field value.

Physics
1 answer:
jonny [76]3 years ago
6 0

Answer:

The electric field value is 240 N/C

Explanation:

Given that,

Distance = 5.0 mm

Potential difference = 1.2 V

We need to calculate the electric field value

Using formula of potential difference

\Delta V=Ed

E=\dfrac{\Delta V}{d}

Where, E = electric field

V = potential difference

d = distance

Put the value into the formula

E=\dfrac{1.2}{5.0\times10^{-3}}

E= 240\ N/C

Hence, The electric field value is 240 N/C

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