Question

The electric field in a region of space has the components Ey = Ez = 0 and Ex = (4.00 N/C)x. Point
A is on the y-axis at y = 3.0 m and point B is on the x-axis at x = 4.0 m. What is the potentia
difference VB - VA?​

Answer 1

Given: The electric field in a region of space has the components Ey = Ez = 0 and Ex = (4.00 N/C) DA.

To find: What is the potential difference between VB - VA?​

Solution:

• Now we have given the electric field in a region of space has the components Ey = Ez = 0 and Ex = (4.00 N/C) DA.

• A is on the y-axis at y = 3.0 m and point B is on the x-axis at x = 4.0 m.

• Now VB - VA = -∫E(x) dx

              VB - VA = -∫4x dx       (lower limit is 0 and upper limit is 4)

              VB - VA = -(4x²/2)       (lower limit is 0 and upper limit is 4)

              VB - VA = -2x²            (lower limit is 0 and upper limit is 4)

              VB - VA = -2(4)² - (-(2(0))        

              VB - VA = -32 - 0

              VB - VA = -32 V

Answer:

            So the potential difference is -32 V.

Answer 2

Given that,

The electric field in a region of space has the components $E_y=E_z=0, E_x=(4\ N/C)x$.

A is on the y-axis at y = 3.0 m and point B is on the x-axis at x = 4.0 m

To find,

$V_B-V_A$

Solution,

The electric field in y and z direction is equal to 0. It only exists in z direction. So, the electric potential in x direction is given by integrating it from 0 to 4 such that :

$V_B-V_A=-\int\limits^4_0 {E_x} \, dx \\\\V_B-V_A=-4\int\limits^4_0 {x} \, dx\\\\V_B-V_A=-4\times [\dfrac{x^2}{2}]^4_0\\\\V_B-V_A=-4\times 8\\\\V_B-V_A=-32\ V$

So, the value of $V_B-V_A$ is (-32 V).