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

Thermodynamics

Space Physics

Solid State Physics

Magnetism

Measurements

Classical Mechanics

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 $V_{B} - V_{A}$ ?