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What is the potential at the centre of a spherical shell ofradius Rthat has a charge Quniformly distributed over its surface? Don’t do an integral of the electric field; just add up the contributions to the potential by the charges.

Short Answer

Expert verified

The potential at the centre of a spherical shell is, V=14πε0QR.

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • Charge, Q
  • Radius, R
  • Small charges, dQ
  • Potential energy, V
02

Understanding Electric Potential:

Electric potential is defined as the effort required to shift a unit charge from one point to another while battling the effects of an electric field. The energy gained by an object when it is moved against an electric field is known as the electric potential energy. Divide the potential energy by the amount of charge to get the charge's electric potential.

03

Determination of the potential at the centre of a spherical shell.

In this problem we calculate the electric potential in the centre of a spherical shell of radiusRand chargeQ. Since the electric field inside the shell is zero, the potential at the centre is the same as the potential on the surface of the sphere. We can consider the sphere as a sum of infinitely many small chargesdQthat are at a distanceRfrom the centre. Each charge generates a potential

dV=14πε0dQR

so, the total potential is

V=∑dvV=∑14πε0dQRV=14πε0QR

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Most popular questions from this chapter

For a path starting at B and going to A (Figure 16.9), calculate (a) the change in electric potential, (b) the potential energy change for the system when a proton moves from B to A, and (c) the potential energy change for the system when an electron moves from B to A. For a path starting at B and going to C, calculate (d) the change in electric potential, (e) the potential energy change for the system when a proton moves from B to C, and (f) the potential energy change for the system when an electron moves from B to C.

Locations A, B and C are in a region of uniform electric field, as shown in Figure 16.66. Location A is at (-0.3,0,0)m. Location B is at (0.4,0,0)m. In the region the electric field has the value (850,400,0)N/C . For a path starting at A and ending at B calculate:

(a) the displacement vector â–³I-.

(b) the change in electric potential,

(c) the potential energy change for the system when a proton moves from A to B ,

(d) the potential energy change for the system when an electron moves from A to B.

LocationsA=<a,0,0>andB=<b,0,0>are on the +x axis, as shown in Figure 16.61. Four possible expressions for the electric field along the x axis are given below. For each expression for the electric field, select the correct expression (1–8) for the potential differenceVA-VB. In each case K is a numerical constant with appropriate units.

(a)E→=<Kx2,0,0>(b)E→=<Kx3,0,0>(c)E→=<Kx,0,0>(b)E→=<Kx,0,0>(1)VA-VB=0(2)VA-VB=K(a-b)(3)VA-VB=K(1a-1b)(4)VA-VB=K(1a3a-1b3b)(5)VA-VB=12K(b2-a2)(6)VA-VB=KIn(ba)(7)VA-VB=K(a3-b3)(8)VA-VB=12K(1a2-1b2)

As shown in Figure 16.72, three large, thin, uniformly charged plates are arranged so that there are two adjacent regions of uniform electric field. The origin is at the center of the central plate. Location A is <-0.4,0,0>m, and location B is<0.2,0,0>m . The electric fieldE1→ has the value <725,0,0>V/m, and E2→is <-425,0,0>V/m.

(d) What is the minimum kinetic energy the electron must have at location A in order to ensure that it reaches location B?

Question: At a particular instant an electron is traveling with speed (6000 m/s) . (a) What is the kinetic energy of the electron? (b) If a proton were traveling at the same speed (6000 m/s), what would be the kinetic energy of the proton?

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