Question

A uniform electric field of magnitude 6.8 × 10 5 N/C points in the positive x direction. (a) Find the electric potential difference (EPD) between the origin (0, 0) and the points (i) (0, 6.0 m), (ii) (6.0 m, 0), and (iii) (6.0m, 6.0 m).

Answer 1

(i) The electric potential difference is 4.08 x 10⁶ V.

(ii) The electric potential difference is 4.08 x 10⁶ V.

(iii) The electric potential difference is 5.77 x 10⁸ V.

How to calculate the electric potential?

The electric potential difference (EPD) between the origin (0, 0) and the points is calculated as;

V = Ed

where;

• E is the electric field strength
• d is the distance

(i) between (0, 0) and (0, 6.0 m),

d = 6 m

V = 6.8 x 10⁵ N/C x 6 m

V = 4.08 x 10⁶ V

(ii) between (0, 0) and (6.0, 0 m),

d = 6 m

V = 6.8 x 10⁵ N/C x 6 m

V = 4.08 x 10⁶ V

(iii) between (0, 0) and (6.0, 6.0 m),

d = √(6² + 6²)

d = 8.485 m

V = 6.8 x 10⁵ N/C x 8.485 m

V = 5.77 x 10⁸ V

Answer 2

i)4080000V

ii)4080000V

iii)5766400v

Step-by-step explanation:

Hello!

To solve this problem we will use the following steps

1. We will use the equation that defines the distance between two points in order to find the distance from the origin to each of the points that the problem asks for.

`d=√(x^2+y^2)`

x= horizonalt component

y=vertical component

d=distance

i)

x=0

y=6

`d=√(0^2+6^2)=6m`

ii)

x=6

y=0

`d=√(6^2+0^2)=6m`

iii).

x=6

y=6

`d=√(6^2+6^2)=8.48m`

2.we calculate the voltage at each point using the equation that relates the distance the voltage and the electric field

V=Ed

where

V=EPD= the electric potential difference

E=electric field magnitude=6.8 × 10^5 N/C.

d=distance

i)

V=Ed=(6.8 × 105 N/C.)(6)=4080000V

ii)

i)

V=Ed=(6.8 × 105 N/C.)(6)=4080000V

iii)

i)

V=Ed=(6.8 × 105 N/C.)(8.48m)=5766400v