In a certain region of space the electric field is uniform and given by E→ = (5.00×102V/m)i^. If the electric potential at the point x = 0, y = 0, z = 0 is equal to V0, find the potential difference V0 − VP for the following point P: x = +5.00 cm, y = 0, z = 0. (a) x = +5.00 cm, y = 0, z = 0; (b) x = +3.00 cm, y = +4.00 cm, z = 0; (c) x = 0, y = +5.00 cm, z = 0; (d) x = -5.00 cm, y = 0), z = 0. a. +25.0 V
b. +15.0 V
c. 0

(a)  25 V

(b)  15 V

(c)  0 V

(d)  -25 V

Explanation:

Notice that the electric field is completely in the x direction (i hat in it expression) so there is only one component of the electric field which is Ex = 500 V/m. We take V0 (at x = 0, y = 0 and z = 0) to be our reference potential.

Recall that in a uniform field the potential (V) is defined as the product of the electric field (E) times the distance (d):

V = E * d

then we use the different values of d (converting them to meters first) to estimate the potential difference at different locations:

(a)  at x = +5.00 cm, y = 0, z = 0:

V = Ex * 0.05 = 500 * 0.05 = 25 V

(b) at x = +3.00 cm, y = +4.00 cm, z = 0:

V = Ex * 0.03 = 500 * 0.03 = 15 V (Notice that Ey and Ez are zero, so they don't contribute)

(c)  at x = 0, y = +5.00 cm, z = 0 (notice that for the x component of the field, the displacement is 0, giving V = 0 V, and also the y and z components of the field are zero, so they also render zero for the potential.

(d)  at x = -5.00 cm, y = 0, z = 0; (again Ey and Ez don't bring any contribution), and V = Ex * (-0.05) = - 25 V

volts times amps 6 x 0.5 3w

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach