Source Charge (Q) +4.31x10³ C 2.33 x10-¹3 C Solutions:. kQ T Distance from Source Charge (r) 7.11 x10-¹2 m 6.45 x10-¹5 m Equation: VE Electric Potential (

A point charge of -8.2 μC is at the origin.What is the electric potential at (3.0 m, - 3.0 m)?

8. Figure below shows a ring of outer radius R = 13.0 cm and inner radius l'inner = 0.200R. It has uniform surface charge density 0 = 6.20 pC/m². With V = 0 at infinity, find the electric potential at point P on the central axis of the ring, at distance z = 2.00R from the center of the ring. 6 dQ K √ ₁7 - Pl What is your dQ? What is your infinitesimal area element? (a) Start with the formula for the potential: V = k What are your vectors r and r'? What is the distance to point P? What is dV? Potential due to a small ring of charge on the disk? (b) Write out the integral that you need to compute to get V. What are the bounds? (c) Once you get an expression for V, solve numerically. (d) Check to see if the units of your expression makes sense for V.

The electric field of ?(?) = 1/r + 3(v/m), where r is the distance from the origin, is applied in a region of space. Find the electric potential between the two points ?1=0,5 m  and r2 =2 m . Hint: you will have to use integration here, with r1 and r2 as your bounds of integration