You have covered a grounded metal surface with a layer of photoconductor. working in the dark, you sprinkle negative charge onto this surface. if you now expose only the left half of the photoconductor to light, you will find that
(a) the left half becomes neutral while the right half remains negatively charged.
(b) nothing happens because there is no changing magnetic field.
(c) negative charge flows from the right side of the photoconductor to the left and both sides become neutral.
(d) the right half becomes neutral while the left half remains negatively charged.

Light from the sun reaches the earth in 8.3 minutes. the velocity is 3.00 x 10^8 m/s. how far is earth from the sun? i know how to get to d = (3 x 10^8 m/s)(498 sec). i just don't know how to get from there to the answer being 1.5 x 10^11 m.

An isolated conducting spherical shell carries a positive charge. part a which statement (or statements) about the electric field and the electric potential inside and outside the spherical shell is correct? which statement (or statements) about the electric field and the electric potential inside and outside the spherical shell is correct? electric potential inside the shell is constant and outside the shell is changing as 1/r2 both the electric potential and the electric field does change with r inside and outside the spherical shell electric potential inside and outside the shell is constant, but not zero electric potential inside the shell is constant and outside the shell is equal to zero electric field inside and outside the shell is constant (does not change with the position r), but is not equal to zero electric field inside and outside the shell is changing as 1/r (where r is the distance from the center of the sphere) electric field inside is equal to zero and outside the shell is constant, but not zero electric potential inside the shell is constant and outside the shell is changing as 1/r electric field inside and outside the shell is changing as 1/r2 electric field inside is equal to zero and outside the shell is changing as 1/r2 electric field inside and outside the shell is zero electric field inside is constant and outside the shell is changing as 1/r

Aleaky 10-kg bucket is lifted from the ground to a height of 15 m at a constant speed with a rope that weighs 0.7 kg/m. initially the bucket contains 45 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 15-m level. find the work done. (use 9.8 m/s2 for g.) show how to approximate the required work by a riemann sum. (let x be the height in meters above the ground. enter xi* as xi.) lim n → ∞ n (9.8)(45−3 xi) i = 1 δx express the work as an integral. 0 dx evaluate the integral. (round your answer to the nearest integer.) j