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Why is the potential difference across the ends of a conductor equal to the product of the electric field and length of the conductor?

Explain this considering I am in high school.

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  • $\begingroup$ Do you know at least the basics of integral calculus? $\endgroup$
    – The Photon
    Jul 10, 2020 at 16:01
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    $\begingroup$ Does this answer your question? How is electric field strength related to potential difference? $\endgroup$
    – Bhavay
    Jul 10, 2020 at 16:06
  • $\begingroup$ Close-voters: I don't think the proposed duplicate is suitable for a high-school-level answer, since it involves multivariable calculus. $\endgroup$ Jul 27, 2020 at 15:15

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This basically comes down to the definition of potential difference.

Potential difference is the energy it would take to move a test charge from one point to another.

And you probably know that the work done when moving something is equal to the force you exerted on it, times the distance you moved it.

And the electric field tells you how much force it would take to move a test charge in a region of space.

In symbols

$$W = F\ell$$

where $W$ is work done, $F$ is force, and $\ell$ is the distance you moved it. Then, for an electric charge moving in an electric field:

$$W = Eq\ell$$

Since the potential energy you added to the system is equal to the work you did on it, you can express the definition of potential difference as

$$V = \frac{W}{q}$$

so

$$V = E\ell$$

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  • $\begingroup$ why is the electric field constant everywhere in the conductor? $\endgroup$ Aug 12, 2020 at 12:33
  • $\begingroup$ i know integral calculus, show it that way. $\endgroup$ Aug 12, 2020 at 13:02

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