Question

A -8.50 nC point charge and a +17.0nC point charge are 18.0 cm apart on the x-axis.

a) What is the electric potential at the point on the x-axis where the electric field is zero?

b) What is the magnitude of the electric field at the point on the x-axis, between the charges, where the electric potential is zero?

Answer 1

Concept - at the point where electric field is zero, the magnitudes of electric field due to both charges must be equal for the field to be zero because their directions are opposite at that point. After finding the position of the point we use potential due to charge to find the potential due to each charge and add them algebraically,

d . 18cm a) So, P -8.5nc 17MC Ep =0 at P ELI = 1E2) or, 1-8.5 nCL = 117 nC) 456o da 47 Go (18+d)2 or, d = 43.46 cm = 0.4346 m So, Up = -8.5 nC I 17MC 45God ' un to (0.18 mt d) = 166.7) v b) V o -8.500 17 nc 4ñ to d 'Art (18-d). -8.5 nc n opy Fate the an d = 6 cm = 0.06m. so Ep = (-8.5 nC) + (17 nC). - 4 Go d2 . Anto (0-18-dj2 = (31875) NIC

***********************************************************************************************
This concludes the answers. If there is any mistake or omission, let me know immediately and I will fix it....