(a) If we place a sixth charged ball at the exact center of the star, with a cha
ID: 2271605 • Letter: #
Question
(a) If we place a sixth charged ball at the exact center of the star, with a charge of +2Q, what is the magnitude of the net electrostatic force exerted on it by the five charged balls shown above?
As shown below, we now replace one of the balls of charge Q (the one that was at the top point of the star) by a ball of charge
Once again, we want to know the magnitude of the net electrostatic force exerted on a ball of charge +2Q, located at the exact center of the star, by the five charged balls shown below. One method to do this is to work out the five individual forces, and add them as vectors, but we should be able to do less work by trying another method, as outlined in parts (b) and (c).
(b) A simpler method is to make use of the answer to part (a). What do you add to the first picture to get the second picture? (Enter a sign and number only.)
The second picture above, with the ball of 3Q, can be thought of as the first picture (the one with five identical charged balls) plus a single charged ball, added at the top point of the star, that has a charge of?
(c) Looked at from the perspective described in (b), the net force on a ball of charge +2Q, located at the exact center of the star in the second picture, is the vector sum of the answer to part (a), plus the force exerted on the ball of charge +2Q by the ball that is the answer to part (b)
This gives a net force, acting on a ball of charge +2Q in the exact center of the star, that has a magnitude of?
The five-pointed star below is a little different, having a ball with a positive charge Q at each point. UseQ = 3.90 10-6 C, and assume that each ball is a distance of 60.0 cm away from the exact center of the star. Also, use k = 9.00 109 N middot m2/C2.Explanation / Answer
part a) From symmetry if you place equal charge at each of the conrners of any regular polygon ,then the electric force exerted by those charges at the centre will be zero.
since in the given figure if you join all the chrges they form a regular pentagon. Hence we can say that electric field at the centr of pentagon is 0.
part b&c) Now for the second case , at the place of -3Q charge assume a combination of charges
i.e. a +1Q charge and -4Q charge . So on the whole the net charge is -3Q.
Now again the +1Q charge and the remaining charges create a senario similar to the first case so the electric field due to these charges at centre will be zero.
Only chrge left is -4Q charge. Force due to this -4Q charge on the +2Q charge at centre is given by
F = kq1q2 / r^2 = 9 x 10^9 x (-4Q) x (2Q) /(0.6 m)^2
= 9 x 10^9 x (-4 x 3.9 x 10^(-6) ) x (2 x 3.9 x 10^(-6)) /(0.36)
= -3.042 N -ve sign indicates the force is away from +2Q charge i.e attractive.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.