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3 years ago

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Assume that charge −q is placed on the top plate, and +q is placed on the bottom plate of a parallel plate capacitor. What is th

e magnitude of the electric field E between the plates? Express E in terms of the variables q, A, and d. Combine all numerical values together include a single multiplier.

1 answer:

Sergeeva-Olga [200]3 years ago

8 0

Answer:

E = $\frac{A(-q)}{d²}$

Explanation:

electric field magnitude, E is the intensity of electric force, F per charge, q applied.

thus, E = force per charge = F/q

there are two charges mention in the question. the charge -q placed on top of the plate is the source charge, while the charge, +q place at the bottom plate is the test charge. two charges is needed for an electric force to occur but for electric field calculation, the test charge, +q is used.

the application of coulomb law is used to determined electric force,

F=[ A *(+q)*(-q)]/d²

where,

A is the coulomb law constant = 9.0 x 10⁹ Nm²/C²

+q = test charge on bottom plate

-q = source charge

d = distance of separation between the top and bottom charge

applying the above equation in E

E = [ A *(+q)*(-q)]/d² DIVIDE q

E = [A(-q)]/d²

E = $\frac{A(-q)}{d²}$

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2 years ago

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2 years ago

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2 years ago

Read 2 more answers

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$f_n = nf_1$

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2 years ago