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Chapter 16: Q6Q (page 662)

In Figure 16.58, what is the direction of the electric field? Is ∆V = Vf −Vi positive or negative?

Short Answer

Expert verified

The direction of the electric field is towards the ball and the potential difference is negative.

Step by step solution

01

Concept/Significance of the potential difference.

A potential difference is the amount of work required to move a charge from one location in the electric field to another.

It refers to the amount of labour or energy necessary to move an electron in a circuit or a specific spatial area of the electrical field.

02

Determination of the direction of the electric field and sign of the potential difference.

The electric field always points from the positive charge to the negative charge. As shown in figure 16.58, thecharges on the ball are negative. it means the electric field is towards the ball.

The positive side has higher potential energy than the negative side. So, the potential difference which is given by,

∆V=Vf-Viis negative as the potential is directed to the negative charges on the ball.

Thus, the direction of the electric field is towards the ball, and the potential difference is negative.

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Most popular questions from this chapter

A capacitor consists of two charged disks of radiusR separated by a distances, whereR>>s. The magnitude of the charge on each disk is Q. Consider points A, B, C, and D inside the capacitor, as shown in Figure 16.88. (a) Show thatΔV=VC−VAis the same for these paths by evaluating ∆V along each path: (1) Path 1:A = B = C, (2) Path 2:A=C, (3) Path 3:A=D=B=C. (b) If,Q=43 μC,R=4 m,s1=1.5 mm ands2=0.7 mm, what is the value ofΔV=VC−VA? (c) Choose two different paths from point A back to point A again, and show that∆V=0for a round trip along both of these paths.

A proton moves from location A to location B in a region of uniform electric field, as shown in Figure 16.5. (a) If the magnitude of the electric field inside the capacitor in Figure 16.5 is 3500 N/C, and the distance between location A and location B is 3 mm, what is the change in electric potential energy of the system (proton + plates) during this process? (b) What is the change in the kinetic energy of the proton during this process? (c) If the proton is initially at rest, what is its speed when it reaches location B? (d) How do the answers to (a) and (b) change if the proton is replaced by an electron?

Question: In a circuit there is a copper wire 40 cm long with a potential difference from one end to the other end of . What is the magnitude of electric field inside the wire?

Locations A, B , and C are in a region of uniform electric field, as shown in Figure 16.65. Location A is at (-0.5,0,0)m. Location Bis at (-0.5,0,0)m . In the region the electric field has the value (750,0,0)N/C. (a)For a path starting at Band ending at C, calculate: (1) the displacement vector△I→ , (2) the change in electric potential, (3) the potential energy change for the system when a proton moves from B to C, (4) the potential energy change for the system when an electron moves from Bto C, (b) Which of the following statements are true in this situation? Choose all that are correct. (1) the potential difference cannot be Choose zero because the electric field is not zero along this path, (2) when a proton moves along this path, the electric force does zero network on the proton, (3) △I→ is perpendicular to E→.

An isolated large plate capacitor (not connected to anything) originally has a potential difference of 1000 V with an air gap of 2 mm. Then a plastic slab 1 mm thick, with dielectric constant 5, is inserted in the middle of the air gap as shown in figure 16.96. Calculate the following potential differences and explain your work.

V1-V2=?V2-V3=?V3-V4=?V1-V4=?
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