2. OV3 points | Previous Answers My Notes Ask Your Teach Two small identical con

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2. OV3 points | Previous Answers My Notes Ask Your Teach Two small identical conducting balls start with different amounts of charge on them. Initially, when they are separated by 85.0 cm (this is the center- to-center distance), one ball exerts an attractive force of 2.40 N on the second ball. The balls are then touched together briefly (charge is conserved in this process), and then again separated by 85.0 cm. Now both balls have a positive charge, and the force that one ball exerts on the other is a repulsive force of 1.10 N. For this problem, use k=9.0×109 Nm2/C2, and assume that the radius of each ball is small compared to 85.0 cm. (a) After the two balls have been touched together, what is the charge on each ball? 0.0000104X C (b) Before they were touched together, one ball had a positive charge. How much charge did that ball have? 0.0000286x C (c) Before they were touched together, one ball had a negative charge. How much charge did that ball have? (Don't forget to include the minus sign!) 0 00079 x C Additional Materials Section 16-3: Coulomb's Law eBook Viewing Saved Work Revert to Last Response

Explanation / Answer

Initially, F= 2.4N

Or, -2.4 = k*q1*q2 / (separation)^2 = 9*10^9*q1*q2/0.85^2 {-ve sign before force is attractive}

Or, q1*q2= -1.926*10^-10 C^2 ____________(1)

Now, charge becomes same on both = q. Since final charge is positive, magnitude of q1> magnitude of q2; where q1>0 and q2<0

Also, charge is conserved so Q= q1+q2 . This charge Q is now divided into two parts and both identical conducting balls now have q= Q/2 each_____________(2)

Now, afetr they have touched each other,

F= k*q*q/0.85^2

Or, 1.1= 9*10^9*q^2/0.85^2

Or, q^2= 8.83*10^-11

So,(a) After the two balls have been touched together, charge on each ball is q= 9.4*10^-6C

Also, Q= 2q= 2*9.4*10^-6C =1.88^10^-5C

from (2); 1.88^10^-5C= q1+q2 and from (1) q1*q2= -1.926*10^-10 C^2

Now, for finding q1 and q2;

So, using (q1+q2)^2=q1^2+q2^2+2*q1*q2;

Or, 7.38*10^-10= q1^2+q2^2 ____________(3)

Also, (q1-q2)^2= q1^2+q2^2-2q1q2

Or, q1-q2= (q1^2+q2^2-2q1q2)^1/2

using q1^2+q2^2 from (3)

Or, q1-q2= ( (7.38*10^-10) - ( 2* (-1.926*10^-10) ) )^1/2 = 3.35*10^-5 ____(4)

Now, adding (q1+q2) with (q1-q2);

2q1= 1.88^10^-5+ 3.35*10^-5

So, (b) Before they were touched together, the ball with the positive charge have q1= 2.615*10^-5C

putting q1 in (4);

Or, 2.615*10^-5 - q2 = 3.35*10^-5

So,(c) Before they were touched together, the ball with the negative charge have q2= -0.735*10^-5C

(d) please upvote to appreciate

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