IIT-JEE 2012 · Paper 2 · Physics Q17

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PhysicsMultiple correct+4Question 17 of 60

Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$ , as shown in the figure. Given that $K=\dfrac1{4\pi\varepsilon_0}\dfrac q{L^2}$ , which of the following statement(s) is(are) correct?

Regular hexagon ABCDEF of side L, centre O: charges +2q at A (left), +q at B (bottom left), −q at C (bottom right), −2q at D (right), −q at E (top right), +q at F (top left). PR is the vertical line through O, ST the horizontal line through O.

The electric field at

O

6K

along

OD

The potential at

O

is zero.

The potential at all points on the line

PR

is same.

The potential at all points on the line

ST

is same.

Answer and worked solution

Official answer: A, B, C

Each charge $q$ at distance $L$ gives a field of magnitude $K$ at O.

(A) Opposite vertices carry opposite charges, so each pair's fields add along the line from the + to the − charge:
- A(+2q), D(−2q): $4K$ along OD;
- B(+q), E(−q): $2K$ along OE;
- F(+q), C(−q): $2K$ along OC.
OE and OC make ±60° with OD, so they add to $2\cdot2K\cos60°=2K$ along OD. Total $6K$ along OD. True.
(B) The charges sum to zero and are all at distance $L$ : $V_O=0$ . True.
(C) PR is the perpendicular bisector of each pair (A, D), (B, C) and (F, E), which carry equal and opposite charges. Every point on it is equidistant from the members of each pair, so $V=0$ all along PR. True.
(D) Along ST (the line through A and D) the potential changes, for example it is large and positive near A. False.

(A), (B), (C)

Topics:Electric field Electric potential Coulomb's law and superposition

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IIT-JEE 2012 · Paper 2, Physics question 17