Solve this Physics Sum ..
[ Grade 12 Physics ]
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Answers
Answer:
Explanation:
charge density=q/4(3.14)r=Q/4(3.14)R.
ie, q=Q*r*r/R*R.
potential at center of shell=kq/r+kQ/R
kQ/R(r/R+1)
Q=charge density*4(3.14)R*R
on simpilfying,
V=charge density*(r+R)/epsilon
or
By superposition princpiple, potential at the common centre is equal to algebraic sum of potentials at centre due to each sphere.
If we want the potential of a sphere, we need the radius (given) and the charge on it (which is what we should find now).
If the total charge is Q, then let’s assume charge of small sphere si q1, and large sphere is q2.
Thus Q = q1 + q2
It is given that the surface charge density is the same, thus:
(q1)/(4*pi*r^2) = (q2)/(4*pi*R^2).
Therefore,
q1 = (r^2)(q2)/(R^2)
But q1 + q2 = Q,
therefore,
q2 = Q(R^2)/(r^2 + R^2),
and similarly (from the same equation,
q1 = Q(r^2)/(r^2 + R^2).
Potential at common centre is now given as:
k(q1)/r + k(q2)/R.
Substituting previously found values, this becomes:
k(Q)(r+R)/(r^2 + R^2).
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hi
charge density=q/4(3.14)r=Q/4(3.14)R.
ie, q=Q*r*r/R*R.
potential at center of shell=kq/r+kQ/R
kQ/R(r/R+1)
Q=charge density*4(3.14)R*R
on simpilfying,
V=charge density*(r+R)/epsilon