Question

The potential in a region between x 0 andx 6.00 m is V a t bx, where a 14.2 V and b-7.70 V/m. (a) Determine the potential at x = 0. Determine the potential at x 3.00 m Determine the potential at x 6.00 m. (b) Determine the magnitude and direction of the electric field at x = 0. magnitude direction Select V/m Determine the magnitude and direction of the electric field at x 3.00 m magnitude V/m direction Select Determine the magnitude and direction of the electric field at x = 6.00 m. magnitude directionSelect V/m

Answer 1

A)1) The potential at x=0 can be given by v(x) at x=0

Therefore, v(0)=a+b(0)= a = 14.2V

2) at x=3 m

V(3)= a+b(3) = 14.2-7.70(3) = -8.9 V

3) at x=6m

V(6) =a+b(6) = 14.27 - 7.70(6) = -32v

B) The electric field E at a position x in terms of potential V can be given by

E = -$\bigtriangledown$V

i.e. -ve gradient of V with respect to x here for one dimension. Therefore

E= -dV/dx = -d(a+bx)/dx = b = 7.70 N-C

For the direction of the electric field, the electric field E points in the direction opposite to the direction of increasing scalar potential φ.( Positive to negative)

Here we can see that the potential is decreasing towards +very x axis. Since, v=a+bx and b is negative. Therefore E points towards+x axis.

For 1) 2) and 3) electric field E is constant therefore

, E= 7.70 V/m.

Direction - towards +x axis

For all the 3 parts