PHYSICS 108 – PROBLEM SET # 1Solutions

Please read Chapter 25 “Electric Forces and Charges” in Resnick, Halliday and Krane Physics, vol 2, 5th Ed.

1. Shortie: We saw in class that a charged comb attracts bits of tissue paper, and after touchingthe comb, the bits of paper often jump away violently. Explain (this is a shortie, youranswer should take up no more than 2 inches of space!)

Rubbing the comb thru your hair gives it a net charge, via friction. The bits of tissue paperare initially electrically neutral, and become polarized (remaining neutral) as the comb ismoved near them, and are therefore attracted to the comb. But sometimes, soon as the tissuepapers touch the comb, if the contact is good enough, the comb then shares its charge withthe tissue, and since both the comb and the tissues are now like-charged, - they repel eachother and fly away. (To further explore this effect, check out the PHET simulation I showed briefly in class:“Balloons and Static Electricity”, linked in our VISTA page: rub the balloon on a sweater tocharge the balloon, then see how the charged balloon interacts with the neutral wall –you canactually “see” the charges.).

2. An electron is in a vacuum near the surface of the Earth. Where should you place a secondelectron so that the net force on the first electron (due to the other electron and gravity) iszero?

Set the magnitudes of the electric and gravitational forces acting upon the electron to be equal:

14πε0

q2

r2= mg Then solve for r:

r =

14πε0

⋅q2

me ⋅ g=

(8.99 ×109N m2 /C 2 ) ⋅ (1.602 ×1019C)2

(9.11×10−31kg) ⋅ (9.81m / s2 )= 5.1m

So the 2nd electron must be placed 5.1 meters below the first electron, along the line joining the first electronand the earth’s center.

(This gives you a feel for the relative strengths of the forces: one little electron 5 meters away from another electron exerts the same force on it as the force of gravity – which is due to the whole earth!)

3. What is the force Fur

on the 1nC charge shown below? (Remember that Fur

is a vector)

Label the charges as shown. The net electric force on q1 is the vector sumof the forces exerted by the other three charges: q2, q3, and q4:

rF2 on 1 =

K q1 q2

r2 , away from q2

=9.0 ×109 N m2 /C2( ) 1×10−9 C( ) 2 ×10−9 C( )

5.0 ×10−2 m( )2 , away from q2

= 0.72 ×10−5 N, away from q2( ) = 0.72 ×10−5 N( ) − cos 45°i − sin 45° j( )

rF3 on 1 =

K q1 q3

r2 , toward q3

=9.0 ×109 N m2 /C2( ) 1×10−9 C( ) 6 ×10−9 C( )

5.0 ×10−2 m( )2 , toward q3

= 2.16 ×10−5 N, away from q3( ) = 2.16 ×10−5 j N

rF4 on 1 =

K q1 q4

r2 , away from q4

= 0.72 ×10−5 N( ) cos 45°i − sin 45° j( )

q1

q2

q3

q4

x

y

Note, you may have noticed that due to the symmetric positioning, and equal sign and magnitudes of chargesof q2 and q4, the x-components of the forces exerted by q2 and q4 cancel – in which case, we can continue withonly examining the y-components of the forces. Either way:

4. Find the force on a positive point charge q located a distance x from the end of a rod oflength L with uniformly distributed positive charge Q (shown below)

rF = F = d∫ F =

14πε0

qdQr2

y=0

y=L

∫ =1

4πε0

q (Q / L)dy(x + L − y)2

y=0

y=L

∫ =qQ

4πε0Ldy

(x + L − y)2y=0

y=L

∫ =qQ

4πε0L1

x + L − y

y=0

y=L

=qQ

4πε0L1x−

1x + L

rF =

qQ4πε0L

1x−

1x + L

, in direction to the right

or qQ4πε0L

x + Lx(x + L)

−x

x(x + L)

=

qQ4πε0L

Lx(x + L)

, to the right

Insurance check: far away from the rod (x >> L) in the limit, this is qQ4πε0x

2 ,as expected - the rod "looks" like a point charge from far away.

5.

y|y=0

|y=L

y-L

x+(L-y)

|y

dy

Divide the rod into a series of infinitesimal lengths, dy, each with charge dQ=(Q/L) dy. (We’re usingthe variable y here because x is the fixed (constant) distance between the charge and the rod)We see that each infinitesimal length contributes to the net force only the horizontal direction to theright, so we need to sum up these horizontal components only. The other two components of the forceare zero.Take the left edge of the rod to be y=0. With this origin, the right edge of the rod is at y=L, and thecharge q is at y=L+x. The distance between the point charge q and each infinitesimal bit of the rodwill therefore be x+(L-y).

⇒

rFon 1 =

rF2 on 1 +

rF3 on 1 +

rF4 on 1 = 2.16 ×10−5 N( ) − 0.72 ×10−5 N( ) 2sin 45°( ) j = 1.14 ×10−5 j N