The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 10.6 V and b = -4.90 V/m. (a) Determine the potential at x = 0. 10.6 V Determine the potential at x = 3.00 m. V (b) Determine the magnitude and direction of the electric field at x = 0. magnitude V/m direction Determine the magnitude and direction of the electric field at x = 3.00 m. magnitude V/m direction

a)

When x =  0 m, V = 10.6 - 4.90*(0) = 10.6 V

When x =  10 m, V = 10.6 - 4.90*(10) = -38.4 V

When x =  6 m, V = 10.6 - 4.90*(6) = -18.8 V

When x =  3.00 m, V = 10.6 - 4.90*(3.00) = -4.1 V

b)

Explanation:

a) If V = a + bx , and a = 10.6 V and b = -4.90 V/m we have:

When x =  0 m, V = 10.6 - 4.90*(0) = 10.6 V

When x =  10 m, V = 10.6 - 4.90*(10) = -38.4 V

When x =  6 m, V = 10.6 - 4.90*(6) = -18.8 V

When x =  3.00 m, V = 10.6 - 4.90*(3.00) = -4.1 V

b) The electric field is:

In this particular case E depends only of b, it means E is a constant value. Therefore, E = 4.9 V/m when x = 0 m and when x = 3.00 m.

I hope it helps you!

unbalanced force

explanation: