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?

Respuesta :

Answer:

121.142 V/m

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

[tex]E= -\frac{dV}{dx}[/tex]

= [tex]\frac{V_1-V_2}{X_2-X_1}[/tex]

=[tex]\frac{5.2-1.8}{2.8\times10^{-2}}[/tex]

= 121.142 V/m