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3 years ago

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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