for y < 0 for 0< y < d. If that was not your > d €0 The answer is V - (у — d) for result, use this potential next. €0 (OV+ OV3 + V E) for y < 0, 0 < y < d (a) Now compute –VV and d < y. ду dz (b) Should your result agree with the electric field E that you calculated in problem 2? Does it agree? What is the value of the integral f E · dr over a closed path? You need to be specially clear and compelling here to earn the points.
for y < 0 for 0< y < d. If that was not your > d €0 The answer is V - (у — d) for result, use this potential next. €0 (OV+ OV3 + V E) for y < 0, 0 < y < d (a) Now compute –VV and d < y. ду dz (b) Should your result agree with the electric field E that you calculated in problem 2? Does it agree? What is the value of the integral f E · dr over a closed path? You need to be specially clear and compelling here to earn the points.
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Question
![**The answer is**
\[
V =
\begin{cases}
\frac{\sigma}{\epsilon_0} y & \text{for } y < 0 \\
0 & \text{for } 0 < y < d \\
-\frac{\sigma}{\epsilon_0} (y - d) & \text{for } y > d
\end{cases}
\]
If that was *not* your result, use *this* potential next.
**6. (a)** Now compute \(-\nabla V = - \left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)\) for \(y < 0\), \(0 < y < d\), and \(d < y\).
(b) Should your result agree with the electric field \(\mathbf{E}\) that you calculated in problem 2? Does it agree?
**7.** What is the value of the integral \(\oint \mathbf{E} \cdot d\mathbf{r}\) over a closed path? You need to be specially clear and compelling here to earn the points.
**8. (a)** Describe the equipotential surfaces with \(V = -5,000\) Volts. Where are they located?
(b) Is there an equipotential volume? If the answer is “yes,” describe it.
(c) Describe all the equipotential surfaces \(V = V_0\) for \(V_0 < 0\) fixed but arbitrary. Where are they located as a function of \(V_0\)?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F289f06d3-ff79-44a0-9e06-63a38c7e8148%2Fc6e1ee2c-01ee-42e1-b2ae-727d0e9f02ac%2Fl8i4kff_processed.png&w=3840&q=75)
Transcribed Image Text:**The answer is**
\[
V =
\begin{cases}
\frac{\sigma}{\epsilon_0} y & \text{for } y < 0 \\
0 & \text{for } 0 < y < d \\
-\frac{\sigma}{\epsilon_0} (y - d) & \text{for } y > d
\end{cases}
\]
If that was *not* your result, use *this* potential next.
**6. (a)** Now compute \(-\nabla V = - \left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)\) for \(y < 0\), \(0 < y < d\), and \(d < y\).
(b) Should your result agree with the electric field \(\mathbf{E}\) that you calculated in problem 2? Does it agree?
**7.** What is the value of the integral \(\oint \mathbf{E} \cdot d\mathbf{r}\) over a closed path? You need to be specially clear and compelling here to earn the points.
**8. (a)** Describe the equipotential surfaces with \(V = -5,000\) Volts. Where are they located?
(b) Is there an equipotential volume? If the answer is “yes,” describe it.
(c) Describe all the equipotential surfaces \(V = V_0\) for \(V_0 < 0\) fixed but arbitrary. Where are they located as a function of \(V_0\)?
Expert Solution
Step 1
6.(a)
Given:
The potential is given by
Introduction:
The electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.
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