. We have a spherically symmetric system as shown. A point charge Q of -6x10-6 C

Question

. We have a spherically symmetric system as shown. A point charge Q of -6x10-6 C is placed at the center of the system. A non-conducting spherical shell has an inside radius of 50 cm and an outside radius of 1 meter. The non-conducting shell has a positive volume charge density of ? = A/r3 ,where r is in meters and A = 2x10-6 C. Find the magnitude of the electric field at a distance of 25 cm from the center of the system. HERE WHAT I DID: q=integral( p 4[PI] R^2 ) (R=25CM , and inner r=50cm on top of the integral) q=A 4[pi] ln r ] ( R=25cm and inner r=50 cm on top of the last line. you know what i mean.) Then E=( K X q)/(R^2) I AM NOT ABLE TO FIND THE CORRECT . PLEASE SHOW ME WHAT I DID WRONG (HERE THE CORRECT ANSWER:8.64X10^5)

Explanation / Answer

If we take the top left particle (or any other, it doesn't really matter - since all the charges are equal and it is a square then the force on each particle due to the other charges will be the same) then the force exerted on it due to the top right charge is:

F(tr) = k*q^2 / r^2 - to the left ;

where k is 1/(4*Pi*s) I've used s to denote the permittivity of free space, and k is approx = 8.988*10^9Nm^2/c^2. And r = the distance between the charges.

Now due to the bottom left charge it is F(bl) = k*q^2 / r^2 - upwards.

The resultant of these two forces is F(1) = sqrt( F(tr)^2+F(bl)^2 )

F(1) = sqrt( 2(k*q^2 / r^2 )^2) = sqrt(2)*k*q^2 / r^2 - up/left

Now we also have the bottom right charge. If the distance to this charge is R then R = sqrt(r^2+r^2) = sqrt(2)*r

so F(br) = k*q^2 / R^2 = k*q^2 / 2r^2 - up/left
which is the same direction as F(1)
Therefore the overall force on the top teft charge is
F = F(1) + F(br)

F = [sqrt(2)*k*q^2 / r^2] + [k*q^2 / 2r^2]
= [(2sqrt(2)+1)*k*q^2] / 2r^2

This is the same force experienced by all the q-charges due to the other q charges (but the force vectors all point away from the centre).

Now for Q the force it excerts of the top left q must be of equal magnitude but in the opposite direction. Since it is at the centre of the square and since the force vectors due to the other q's are pointing away from the centres we know that Q must be negaitive. Since we know this we can simply find the magnitude of Q by:

Now F(Q) = F
However the distance to Q = 1/2 the distance to the bottom right q; So r(Q) = sqrt(2)*r / 2
k*q*Q / (sqrt(2)*r/2)^2 = [(2sqrt(2)+1)*k*q^2] / 2r^2

Simplifying: 4Q / 2r^2 = (2sqrt(2)+1)*q / 2r^2
Q = [2sqrt(2)+1]*q/4
Q = [2sqrt(2)+1]*(9.29*10^-9)/4 =8.64 X 10^5

However we said Q must be negative, so
Q =8.64 X 10^5

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