Chapter 24: Q93P (page 716)

A uniform charge of $16.0 μ C$ is on a thin circular ring lying in xy plane and centered on the origin. The ring’s radius is $3.00 c m$ . If point A is at the origin and point B is on the z-axis at $z = 4.00 c m$ , what is $V_{B} - V_{A}$ ?

Short Answer

Expert verified

The value of $V_{B} - V_{A}$ is $- 1.92MV$ .

Step by step solution

Step 1: Given data:

The uniform charge, $q = + 16.0 μ C$ ,

The radius of the circle, $R = 0.03 m$

The vertical distance, $A B = Z = 0.04 m$ .

Determining the concept:

Potential at a point due to a charge q is given as,

$V = \frac{1}{4 {πε}_{0}} \frac{q}{r}$

Here, is the distance of the point from the charge, $q$ is the electric charge, and $\frac{1}{4 {πε}_{0}}$ is the Coulomb constanthaving a value $9 \times 10 N . m^{2} / C^{2}$ .

Determining the VB-VA:

Due to the ring, the electric potential at $A$ is,

$V_{A} = \frac{1}{4 π ε_{o}} \cdot \frac{q}{R}$

$= (9.0 \times 10^{9} N . m^{2} / C^{2}) \times \frac{16.0 \times 10^{- 6} C}{(0.03 m)}$

$= 4800 \times 10^{3} V$

$= 4 . 8 MV$

Again, due to the ring, the electric potential at $B$ is,

$V_{B} = \frac{1}{4 π ε_{o}} \cdot \frac{q}{\sqrt{R^{2} + Z^{2}}}$
$V_{B} =$ $(9.0 \times 10^{9} N . m^{2} / C^{2})$ $\times \frac{16 \times 10^{- 6} C}{\sqrt{{0.03}^{2} + {0.04}^{2}} m}$

$= (9.0 \times 10^{9} N . m^{2} / C^{2})$ $\times \frac{16 \times 10^{- 6} C}{0.05 m}$

$= 2,880 \times 10^{3} V = 2.8 8MV$

Now,

$V_{B} - V_{A} = 2.88MV - 4.8MV = - 1.92MV$

Hence, potential at $V_{B} - V_{A}$ is $- 1.92MV$ .

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A uniform charge of $16.0 μ C$ is on a thin circular ring lying in xy plane and centered on the origin. The ring’s radius is $3.00 c m$ . If point A is at the origin and point B is on the z-axis at $z = 4.00 c m$ , what is $V_{B} - V_{A}$ ?

Short Answer

Step by step solution

Step 1: Given data:

Determining the concept:

Determining the VB-VA:

One App. One Place for Learning.

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A uniform charge of 16.0μCis on a thin circular ring lying in xy plane and centered on the origin. The ring’s radius is3.00cm. If point A is at the origin and point B is on the z-axis at z=4.00cm, what isVB-VA?

Short Answer

Step by step solution

Step 1: Given data:

Determining the concept:

Determining the VB-VA:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electromagnetism

Electric Charge Field and Potential

Quantum Physics

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Radiation

Engineering Physics

Study anywhere. Anytime. Across all devices.

A uniform charge of $16.0 μ C$ is on a thin circular ring lying in xy plane and centered on the origin. The ring’s radius is $3.00 c m$ . If point A is at the origin and point B is on the z-axis at $z = 4.00 c m$ , what is $V_{B} - V_{A}$ ?