Please solve Quetion 2 Electric field Consider n equal positively charged partic
ID: 1572344 • Letter: P
Question
Please solve Quetion 2
Electric field
Consider n equal positively charged particles each of magnitude Q/n placed symmetrically around a circel of radius a. Calculate the magnitude of the electric field at a point a distance x from the centre if the circle and on the line passing through the centre and perpendicular to the plane of the circle. Now consider a ring of radius a that carries a uniformly distributed positive total charge Q. Recall from class the calculation of the electric field at point a point a distance x from the centre of the ring and on the line passing through the centre and perpendicular to the plane of the ring. Explain why the result in part (a) is identical to the result for the ring.Explanation / Answer
a) E.F at a distance x passing through center and perpendicular to plane of paper due to n point charges will be equal to E.F due to sum of individual charge.
E.F = E1 + E2 + E3 +--------------------+ En
= [k*( Q/n) / x2 ]+ [k*( Q/n) / x2 ]+ [k*( Q/n) / x2 ]+------------------------------+ [k*( Q/n) / x2 ]
here we are adding n times so,E.F = n* k*( Q/n) / x2
= k*Q/ x2
b)Electric field at a distance x ( x>a) due to ring having a positive charge Q uniformly distributed around it:
E.F = k*Q / x2
[ if x <a that is if point lies inside the ring, E.F= 0]
result of part a is identical to b, because in part a total charge will be Q due to n point charges ( which comes out due to addition of n point charges Q/n + Q/n + ------------ to n times = n*Q/n = Q). and ring also has charge Q .so E.F will be same .also both have same radius and at same point E.F calculated.