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Chapter 25: Q. 30 (page 710)

What is the electric potential at the point indicated with the dot

in FIGURE EX25.30?

Short Answer

Expert verified

The value of electric potential energy at the point indicated with the dot is 1770VV

Step by step solution

01

Definition of  electric potential  energy  

The value of potential charge of the q point is V

V=14πε0qr

Here q is the charge value , r is the distance ϵ0Permittivity of the free space.

Then the total value of potential ∑iVi=14πε0qiri

The magnitude of all charges in Point 1 and 3 is 2.0 nC.

q1=q3=2.0nC10-19nC1nC=2.0×10-19C

Magnitude of all charges is in Point 2 is 4.0 nC

Nowq2=4.0nC10-19nC1nC=4.0×10-19C

02

Point indication with the dot  points 

The value of distance from the unit positive charge of the 2

r2=42+32=5cmV=14πε0q1r1+14πε0q2r2+14πε0q3r314πε0=9×109N.m2

Now substitutingall the value to this equation;

V=9×109N.m2×2×10-19C310-2m1cm+4×10-19C510-2m1cm+2×10-19C410-2m1cm=1770V

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Most popular questions from this chapter

A 5.0-cm-diameter metal ball has a surface charge density of 10mC/m2. How much work is required to remove one electron from this ball?

Your lab assignment for the week is to measure the amount of charge on the 6.0-cm-diameter metal sphere of a Van de Graaff generator. To do so, you’re going to use a spring with a spring constant of 0.65 N/m to launch a small, 1.5 g bead horizontally toward the sphere. You can reliably charge the bead to 2.5 nC, and your plan is to use a video camera to measure the bead’s closest approach to the edge of the sphere as you change the compression of the spring. Your data is as follows:

Use an appropriate graph of the data to determine the sphere’s charge in nC. You can assume that the bead’s motion is entirely horizontal, that the spring is so far away that the bead has no interaction with the sphere as it’s launched, and that the approaching bead does not alter the charge distribution on the sphere.

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.2 cm apart with a 25 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.

a. What is the electric field strength between the plates?

b. With what speed does an electron exit the electron gun if its entry speed is close to zero?

Note: The exit speed is so fast that we really need to use the theory of relativity to compute an accurate value.

Your answer to part b is in the right range but a little too big.

A proton is released from rest at the positive plate of a parallelplate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 50,000 m/s. What will be the final speed of an electron released from rest at the negative plate?

Two point charges qa and qb are located on the x-axis at x = a and x = b. FIGURE EX25.34 is a graph of V, the electric potential.

a. What are the signs of qa and qb?

b. What is the ratio qa/qb?

c. Draw a graph of Ex, the x-component of the electric field, as a function of x.
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