Question

would lke to Show Work for this question Question

Answer 1

Your picture quality is not clear so check this values which I understood from picture: -26e & +46e & +14e. If any of the values are different let me know in comment section, I will recalculate it.

Solution:

Remember whenever metal spheres are touched, the charge is equally distributed between them. now Suppose when two spheres A and B are touched, After separating

Charge on A = Charge on B = (qa + qb)/2

Using this:

When Sphere W and A are touched and separated

Charge on A = Charge on W = (qw + qa)/2

qw = 0, given initially

Suppose qa = Charge on Sphere A = ne

Charge on A = Charge on W = (0 + ne)/2 = ne/2

Now when Sphere W and B are touched

qb = -26e

Charge on W = Charge on B = (ne/2 - 26e)/2 = ne/4 - 13e

Now when Sphere W and C are touched

qc = 46e

Charge on W = Charge on C = ((ne/4 - 13e) + 46e)/2 = ne/8 + 33e/2

Now given that final charge on sphere W = 14e

So,

ne/8 + 33e/2 = 14e

ne/8 = 14e - 33e/2

ne/8 = -5e/2

n/8 = -5/2

n = -20

So Initial charge on Sphere A = -20e = (multiple of -20)*e

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