The electric potential from particle 1 is 35 aVoltsS at the origin. Particle 2 has twice the charge and is twice the distance from the origin as Particle 1. Particle 3 is the same distance from the origin and same charge as particle 1 d 2d - Part A Determine the net electric potential from all three charges at the origin. Input your answer in Volts. VOAE Vt = Volts Submit Request Answer Next>

Electric Potential of Distributed Charge 4. An evenly charged wire of length L has a total charge of Q distributed on it and is located on the x-axis a distance a away from the origin as drawn. Set up the integral that could be solved to determine the electric potential at any point on the y-axis, (0,yo). You do not have to evaluate the integral.

Please write clearly

Attached is question, multiple choice.

The figure(Figure 1) is a graph of E The potential at x=2m is -200 V Part A You may want to review (Pages 715 - 716) What direction is the x-component of the electric field pointing between x=2m and x=3m in the graph? Express your answer to two significant figures and include the appropriate units. E, is pointing rightward (+x-direction) O Not enough information O E, is pointing down (minus y-direction) O E, is pointing up (+y-direction) O E is pointing leftward ( minus x-direction) Figure 1 of 1 Submit Request Answer E, (V/m) Provide Feedback 200 100 x (m) 3. Pearson

tion A1 The electric potential V(x) for a planar charge distribution is as follows: (= (1 + 2)² =a a V(x) = where Vo is the potential at the origin and a is a distance. Derive an expression for the corre- sponding electric field Ē(x).