Question

The distance between the points A and B on two equipotential lines with V1=5.2 V and V2=1.8 V is 2.8 cm. What is the average electric field (in Volt/m) at the midpoint C expressed to one decimal place?

Answer 1

121.142 V/m

Step-by-step explanation:

A and B are on two equipotential line.

Equipotential line are imaginary lines in space along which the potential does not change.

to find electric field at the midpoint C

we use the formula

`E= -(dV)/(dx)`

=
`(V_1-V_2)/(X_2-X_1)`

=
`(5.2-1.8)/(2.8*10^(-2))`

= 121.142 V/m