solutions_26 - PHY2061 R. D. Field Chapter 26 Solutions...

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PHY2061 R. D. Field Solutions Chapter 26 Page 1 of 14 Q1 L/3 L Q2 P Chapter 26 Solutions Problem 1: What is the magnitude ( in milliN ) and direction of the electrostatic force on a -2.0 microC charge in a uniform electric field given by x C N E & ) / 100 ( = r ? Answer: 0.2 in the -x direction Solution: The electrostatic force is given by, x mN x C N C E q F & 2 . 0 & ) / 100 )( 0 . 2 ( = = = µ r r . Problem 2: Two point charges, +8 nanoC and -2 nanoC lie on the x-axis and are separated by 6 meters as shown in the Figure . What is the magnitude of the electric field ( in N/C ) at a point P on the x-axis a distance of 6 meters to the right of the negative charge? Answer: zero Solution: The electric field at the point P is the superposition of the electric fields from each of the two charged as follows: r r rr r E KQ L x E KQ L x KQ L x EE E tot 1 2 2 22 12 4 2 0 =− == =+= $ () $$ , where Q = 2 nC and L = 6 m . Problem 3: Two charges Q 1 and Q 2 are separated by distance L and lie on the x-axis with Q 1 at the origin as shown in the figure . At a point P on the x-axis a distance L/3 from Q 1 the net electric field is zero. What is the ratio Q 1 /Q 2 ? Answer: 0.25 Solution: In order for the net electric field to be zero it must be true that 6 m 6 m -2 nC P +8 nC
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PHY2061 R. D. Field Solutions Chapter 26 Page 2 of 14 2 2 2 1 ) 3 / 2 ( ) 3 / ( L KQ L KQ = , which implies that Q 1 /Q 2 = 1/4 . Problem 4: A particle with a charge to mass ratio of 0.1 C/kg starts from rest in a uniform electric field with magnitude, E = 10 N/C . How far will the particle move ( in m ) in 2 seconds ? Answer: 2 Solution: We know that qE ma F = = so that the acceleration is E m q a = . Thus, the distance traveled is m s C N kg C Et m q at d 2 ) 2 )( / 10 )( / 1 . 0 ( 2 1 2 1 2 1 2 2 2 = = = = Problem 5: A large insulating sheet carries a uniform surface charge of density -1.2 microC/m 2 . A proton (with a mass of 1.67x10 -27 kg ) is released from rest at perpendicular distance of 1.0 m from the sheet. How much time ( in microsec ) elapses before the proton strikes the sheet? Answer: 0.55 Solution: We know that F = eE = Ma so that a = eE/M . We also know that if the particle starts from rest then the distance traveled, d = at 2 /2 . Solving for t gives, s m C C Nm C kg m e dM eE dM a d t µ σ ε 55 . 0 ) / 10 2 . 1 )( 10 6 . 1 ( )) /( 10 85 . 8 )( 10 67 . 1 )( 1 ( 4 4 2 2 2 6 19 2 2 12 27 0 = × × × × = = = = , where I used E = σ /(2 ε 0 ) .
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PHY2061 R. D. Field Solutions Chapter 26 Page 3 of 14 Problem 6: At a distance of 1 meter from an isolated point charge the electric field strength is 100 N/C . At what distance ( in m ) from the charge is the electric field strength equal to 50 N/C ? Answer: 1.41 Solution: We know that E 1 =KQ/r 1 2 and E 2 =KQ/r 2 2 so that E E r r 1 2 2 2 1 2 = or rr E E m NC m 2 2 1 2 1 2 22 1 100 50 2 == = () / / . Hence, rmm 2 21 4 1 . . Problem 7: The magnitude of the electric field 300 m from a point charge Q is equal to 1,000 N/C . What is the charge Q ( in C )?
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solutions_26 - PHY2061 R. D. Field Chapter 26 Solutions...

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