An electrically neutral thin, square metal slab, measuring 5.40cm on a side, is

Q: An electrically neutral thin, square metal slab, measuring 5.40cm on a side, is

direction of the external field = in the direction top to bottom of slab direction of the field that is generated by the charges induced on the slab = bottom to top of slab electric field induced = 1410N/C (equal and opposite in direction) E = sigma/e = (q/(0.054^2 * 8.85*10^-12)) 1410 = (q/(0.054^2 * 8.85*10^-12)) => q = 36.387306 pC

ID: 2280846 • Letter: A

Question

An electrically neutral thin, square metal slab, measuring 5.40cm on a side, is placed in a uniform external field that is perpendicular to its square area.

If the top of the slab becomes negatively charged, what is the direction of the external field?

What is the direction of the field that is generated by the charges induced on the slab?

If the external field strength is 1410N/C , what is the strength of the field that is generated by the charges induced on the slab? (N/C)

What is the total charge on the negative side of the slab? (pC)

Explanation / Answer

direction of the external field = in the direction top to bottom of slab

direction of the field that is generated by the charges induced on the slab = bottom to top of slab

electric field induced = 1410N/C (equal and opposite in direction)

E = sigma/e = (q/(0.054^2 * 8.85*10^-12))

1410 = (q/(0.054^2 * 8.85*10^-12))

=> q = 36.387306 pC

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An electrically neutral thin, square metal slab, measuring 5.40cm on a side, is

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