5 x (cm) 2q 10 x (cm) 0 2q +3q Figure 18.52 (a) Point charges located at 3.00, 8

Question

5 x (cm) 2q 10 x (cm) 0 2q +3q Figure 18.52 (a) Point charges located at 3.00, 8.00, and 11.0 cm along the x-axis. (b) Point charges located at 1.00, 5.00, 8.00, and 14.0 cm along the x-axis 42. (a) Find the total electric field at x- 1.00 cm in Figure 18.52(b) given that q -5.00 nC. (b) Find the total electric field at x = 1 1.00 cm in Figure 18.52(b), (c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, will there be a single charge, double charge, etc., and what will its value(s) be?)

Explanation / Answer

We know that E =kq/r2 and that E fields are vectors. We must therefore determine the directions of the fields before we can perform vector addition. E fields from + charges point radially outwards, from - charges point radially inwards.
(a)

Find the total electric field at x = 1.00cm given that q = 5.00nC.
Let the E fields be E1 from the charge at 1cm, E5 from the charge at 5cm, E8, E14 etc.
For the charge at 1cm, you have r = 0 in the denominator which is infinity which seems rather a pointless exercise. As you approach 1cm the fields from the other charges are swamped by the field from the charge at 1cm because it's denominator is approaching zero:
k(2q)/0.0000001^2 = 9e15 N/C pointing to the left on the right side and ponting to the right on the left side of the charge at 1cm. So yes, right at 1cm, the E field rises to infinity.

(b)

The distances from the charges to 11cm are -2q:10cm, q:6cm, 3q:3cm, -q:3cm
At 11cm E5, E8 and E14 all point towards the right and add. E1 points to the left and subtracts:
E = -2kq/0.10^2 + kq/0.06^2 + 3kq/0.03^2 + kq/0.03^2 = 203,219 N/C
The only way you can get 212,000 is if you DON'T subtract the field from the charge at 1cm so 2.12e5 N/C at 11cm is wrong.

(c)

The -q at 14cm is attracted to the 3q at 8cm and they combine and become 2q. The q at 5cm is attracted to the -2q at 1cm and they combine for -q which combines with the 2q for a final charge of q

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