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Two2.0cmdiameter insulating spheres have a6.0cmspace between them. One sphere is charged to+10nC, the other to-15nC. What is the electric field strength at the midpoint between the two spheres?

Short Answer

Expert verified

The electric field strength at the midpoint| between the two spheres is5.56×105N/C.

Step by step solution

01

Formula for electric field intensity

The intensity of the electric field at a point dunits away from a point charge qcan be calculated using the following formula:

(E)=q/4πεd2NC-1

The vector sum of the intensities produced by the individual charges equals the intensity of the electric field at any point due to a number of charges.

02

Calculation for electric field at the midpoint| between the two spheres 

Distance between the outer edges is 6.7cm.

Diameter of sphere is2.0cm.

Hence the distance between center is(6.7+2.0)=8.7cm.

Mid point is ,

=(8.7/2)cm

localid="1648629844607" =4.35cm=0.0435m.

Charge of spherelocalid="1648629881080" 1islocalid="1648391289846" 41nC.

Charge of spherelocalid="1648629893533" 2is-76nC.

The fields will now all point in the same direction.

As a result, whole field is:

localid="1648630029365" Enet=E1+E2

localid="1648630472057" E1due tolocalid="1648630003851" 42nCwill be:

localid="1648630478720" E1=9×109×41×10-9(0.0435)2

localid="1648630485528" E1=1.95×105N/C

localid="1648630491962" E2due to localid="1648630497375" -76nCwill be:

localid="1648630504401" E2=9×109×76×10-9(0.0435)2

localid="1648630510652" E2=3.61×105N/C

03

Calculation for electric field strength at the midpoint| between the two spheres 

SubstituteE1andE2values we get,

Enet=E1+E2

Enet=1.95×105N/C+3.61×105N/C

Enet=5.56×105N/C

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Most popular questions from this chapter

The electric field strength2.0cmfrom the surface of a10cmdiameter metal ball is50,000N/C. What is the charge (innC)on the ball?

You have a summer intern position with a company that designs and builds nanomachines. An engineer with the company is designing a microscopic oscillator to help keep time, and you’ve been assigned to help him analyze the design. He wants to place a negative charge at the center of a very small, positively charged metal ring. His claim is that the negative charge will undergo simple harmonic motion at a frequency determined by the amount of charge on the ring.

a. Consider a negative charge near the center of a positively charged ring centered on the z-axis. Show that there is a restoring force on the charge if it moves along the z-axisbut stays close to the center of the ring. That is, show there’s a force that tries to keep the charge at z=0. b. Show that for small oscillations, with amplitude <<R, a particle of mass mwith charge-qundergoes simple harmonic motion with frequency f=12πqQ4πε0mR3,RandQare the radius and charge of the ring.

c. Evaluate the oscillation frequency for an electron at the center of a 2.0μmdiameter ring charged to 1.0×10-13C.

In Problems 63 through 66 you are given the equation(s) used to solve a problem. For each of these

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem

(9.0×109Nm2/C2)(2.0×10-9C)s(0.025m)3=1150N/C

The electric field 5.0cmfrom a very long charged wire is ( 2000N/C, toward the wire). What is the charge (in nC) on a 1.0-cm-long segment of the wire?

Air "breaks down" when the electric field strength reaches 3.0×106N/c, causing a spark. A parallel-plate capacitor is made from two 4.0cm×4.0cm electrodes. How many electrons must be transferred from one electrode to the other to create a spark between the electrodes?

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