Question

One point has a voltage of 9 volts and a second point 5 millimeters away has a voltage of 3 volts What is the strength of the electric field between these points?

Answer 1

The strength of the electric field between two points with a voltage difference of 6 volts and separated by 5 millimeters is 1200 V/m.

Step-by-step explanation:

The electric field strength between two points where the voltages are different can be calculated using the formula E = V/d, where E is the electric field strength, V is the voltage difference between the two points, and d is the distance separating them. In this case, the voltage difference is 9 volts - 3 volts = 6 volts, and the distance is 5 millimeters, which is 0.005 meters. To find the electric field strength, we divide the voltage difference by the distance: E = 6 V / 0.005 m = 1200 V/m. Therefore, the strength of the electric field between these points is 1200 V/m.

Answer 2

1,200V/m

Step-by-step explanation:

I will call the first point A and the second point B, we have:

voltage in A:
`V_(A)=9V`

Voltage in B:
`V_(B)=3V`

The distance between the points:
`d= 5mm = 5x10^(-3)m`

And we calculate the electric field due to the difference of potential as follows:

`E=(V_(A)-V_(B))/(d)`

Substituting known values:

`E=(9V-3V)/(5x10^(-3)m)=(6V)/(5x10^(-3)m) = 1,200V/m`

the strength of the electric field between these points 1,200V/m