Question

The electric field in a region of space has the components $E_y$ = $E_z$ = 0 and $E_x$ = (4.00 N/C)ₓ. Point A is on the y axis at y = 3.00 m, and point B is on the x axis at x = 4.00 m. What is the potential difference $V_B - V_A$?

Answer 1

Answer:

the potential difference between point B and A is -32Volts

Explanation:

The electric field is given as

Ex=4x N/C

Ey=0

Ez=0

E=(4x 0 0) N/C

The point A is on x axis at 4m and point B is on y axis at 3m,

V is given as

Vb-Va=-∫ E.ds

Since the field has value only at the x axis and it is parallel to the x axis, then, the angle between the field and the x axis is zero(0)

Therefore, E.ds = EdsCosθ

And θ = 0

Then, E.ds =Eds

And since it is only in the direction of x, then ds=dx

So,

Vb-Va=-∫ E.ds

Vb-Va=-∫ Edx. From x=0 to x=4

Where E= 4x

Vb-Va=-∫ 4xdx. From x=0 to x=4

Vb-Va=-4x² /2. From x=0 to x=4

Vb-Va=-2x² from x=0 to x=4

Vb-Va= -2(4²-0²)

Vb-Va= -2(16-0)

Vb-Va= -32Volts

So, the potential difference between point B and A is -32Volts