JEE Advanced
Physics
Electrostatics
2022
JEE Advanced 2022 (Paper 1)
JEE Advanced Physics Question (2022) — Solution
Question
Six charges are placed around a regular hexagon of side length a as shown in the figure. Five of them have charge q , and the remaining one has charge x . The perpendicular from each charge to the nearest hexagon side passes through the center O of the hexagon and is bisected by the side. Which of the following statement(s) is(are) correct in SI units?
Options
- A. When x = q , the magnitude of the electric field at O is zero.
- B. When x = - q , the magnitude of the electric field at O is q 6 π ∈ 0 a 2 .
- C. When x = 2 q , the potential at O is 7 q 4 3 π ∈ 0 a .
- D. When x = - 3 q , the potential at O is 3 q 4 3 π ϵ 0 a .
Answer
C. When x = 2 q , the potential at O is 7 q 4 3 π ∈ 0 a .
Step-by-step solution
The given arrangements of the charges also form a regular hexagon with side length equal to 3 a When x = q , the situation is symmetric ⇒ Electric field at O would be zero. ⇒ A is correct. When x = - q , we can think of x as q + - 2 q . Therefore, the magnitude of the electric field will be equivalent to that if only charge of - 2 q is kept at the the bottom point. Hence, E O = 1 4 π ϵ 0 2 q 3 a 2 = 1 4 π ϵ 0 2 q 3 a 2 = q 6 π ϵ 0 a 2 ⇒ B is correct For x = 2 q , potential at O is V O = 5 × 1 4 π ϵ 0 q 3 a + 1 4 π ϵ 0 2 q 3 a = 7 q 4 3 π ϵ 0 a ⇒ C is correct For x = - 3 q , V O = 5 × 1 4 π ϵ 0 q 3 a + 1 4 π ϵ 0 - 3 q 3 a = q 2 3 π ϵ 0 a ⇒ D is not correct.
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