A pair of oppositely charged parallel plates are separated by 5.39 mm. A potential difference of 578.0 V exists between the plates. What is

Question

A pair of oppositely charged parallel plates are separated by 5.39 mm. A potential difference of 578.0 V exists between the plates. What is the magnitude of the electric field strength in the region that is located between the plates? The fundamental charge is 1.602 × 10−19 C . Answer in units of V/m.

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Amity 5 years 2021-09-05T10:45:05+00:00 1 Answers 15 views 0

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  1. ThanhThu
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    2021-09-05T10:46:29+00:00
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    Answer:

    E = 1.07*10⁵ V/m

    Explanation:

    • By definition, the electric potential at a given point in space, is the work done by unit charge, by the electrostatic force, in order to move the charge from infinity (assumed as a zero reference potential level) to that point.
    • So, the potential difference, is the work done, per unit charge, to move a charge from one point to another.
    • In the case of opposite charged plates, the electric field (force per unit charge) is normal to the plates and constant through the entire surface of the plates, so it will be parallel to the displacement of a charge under the influence of this field.
    • So, we can find the work done by the field, moving a positive test charge from the positive plate to the other, as follows:

           W = F* d = Q*E*d\\\\ \frac{W}{Q} = \Delta V = E*d

    • We can solve for E, as follows:

           E =\frac{\Delta V}{d} =\frac{578.0 V}{0.00539 m} = 1.07e5 V/m

    • The magnitude of the electric field strength in the region that is located between the plates is 1.07*10⁵ V/m.

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