# Electric Field and Potential: Problem 1.04 & 1.05 with detailed Solutions

### A) 2B) 1C) 3D) None of these

Let five equal positive point charges (say each +q) are placed at the vertices A, B, C, D and E of a regular hexagon ABCDEF. No charge is present at the sixth vertex F.
As the points A, O and D are collinear, AO = OD and the points A & D carry like charges of equal magnitude, the individual fields produced by the charges located at A and D at the centre O will be equal and opposite. Thus, whatever field is produced by the charge +q (present at point A) at the centre O, that is cancelled out by another oppositely directed field at O caused by the charge +q present at point D (Superposition Principle). We conclude that the identical point charges located at any pair of opposite vertices produces zero resultant field at the the centre O.

Thus, net field at point O due to the charges at the vertices A, D, B & E is null. Effectively, the electric field at the centre O is only due to the charge present at the vertex C. Hence, net field at O = Field at O due to charge at C = E (given in the problem).

Now in the next case, an additional point charge -q is introduced in the system at the vertex F. Observe that the field produced by +q (located at C) and the field produced by -q (located at F) at the centre O will be equal in magnitude (both E) and parallel to each other. So both add up to yield a resultant of 2E. Thus E' = 2E or E'/E = 2.

### 1.05) A non-conducting ring of radius R contains uniformly distributed charge +Q. A tiny part of the ring, of length d, is removed (d << R). The net electric field at the centre of the ring will now beA) directed towards the gap, inversely proportional to R³B) directed towards the gap, inversely proportional to R²C) directed away from the gap, inversely proportional to R³D) directed away from the gap, inversely proportional to R²

Electric field produced by a uniformly charged ring at its centre is zero. This is due to the symmetry of the charge distribution. For any small section of the ring (carrying some charge) there exists an identical section (carrying same charge) at the diametrically opposite point. Such two sections nullify each other's field at the centre and therefore the net field at centre is zero.

When a small section of the ring is removed, its corresponding identical section located at the diametrically opposite point is left to cause some resultant field at the centre. This field will be directed towards the gap as the charge is positive. Also, the field magnitude equals k(dq)/R², R being the radius of the ring.

Hence, the field at centre is inversely proportional to R³ and directed towards the gap.