Question

The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 12.6 V and b = -7.90 V/m.

(a) Determine the potential at x = 0.
______ V

Determine the potential at x = 3.00 m.
______ V

Determine the potential at x = 6.00 m.
______ V

(b) Determine the magnitude and direction of the electric field at x = 0.

magnitude	________ V/m
direction	+x or -x?

Determine the magnitude and direction of the electric field at x = 3.00 m.

magnitude	________ V/m
direction	+x or -x?

Determine the magnitude and direction of the electric field at x = 6.00 m.

magnitude	________ V/m
direction	+x or -x?

Answer 1

a) The equation for the potential is given as, V=a+bx where a=12.6 V and b = -7.90 V/m. Thus substituting these values in the equation, We get, V= 12.6 - 7.90 x ......... (1)

(1) Is our equation for potential. Now we have to find out the value of this potential at x=0,3 and 6 m. To do this we will simply put the values of x in the above equation (1).

Thus the potential (at x=0) = (12.6 - 7.90 X 0) V

= 12.6 V

Similarly the potential (at x=3) = (12.6 - 7.90 X 3) V

= 12.6 - 23.7 V
= - 11.1 V

Furthermore the potential (at x=6) = (12.6 - 7.90 X 6) V

= 12.6 - 47.4 V

= -34.8 V

Thus the value of potential at x=0, 3 and 6 m is found out.

b) According to the electrodynamics, Electric field (E) and Electric potential (V) are related by
According to this definition we can find out the equation of electric field (E) by differentiating with respect to x. Thus,

= - b (b= -7.90 V/m)

= 7.90 V/m

Since the equation of Electric field (E) lacks any x component, That shows us that the magnitude and the direction of Electric field remains constant irrespective of the value of x. Thus, E= 7.90 V/m for x=0, 3 and 6 and it is directed towards the +x axis.