Solveeit Logo
Login

Question

Question: Consider a set of 12 equal charges (+q each), arranged to form two concentric regular hexagons in th...

Consider a set of 12 equal charges (+q each), arranged to form two concentric regular hexagons in the xyxy-plane, centred at origin. The edges of the two hexagons have lengths aa and 2a2a respectively. They are arranged in such a way that two vertices of each one of them are on the yy-axis. Suppose that the charges from the vertices at (0,2a)(0,2a) and (0,−a)(0, -a) are removed. Then, the magnitude of the electric field produced by the resulting configuration at the origin is

A

38πϵ0a2q\frac{3}{8\pi\epsilon_0a^2}q

B

316πϵ0a2q\frac{3}{16\pi\epsilon_0a^2}q

C

516πϵ0a2q\frac{5}{16\pi\epsilon_0a^2}q

D

58πϵ0a2q\frac{5}{8\pi\epsilon_0a^2}q

Answer

316πϵ0a2q\frac{3}{16\pi\epsilon_0a^2}q

Explanation

Solution

The problem describes a configuration of 12 equal charges (+q each) arranged to form two concentric regular hexagons in the xyxy-plane, centered at the origin. The inner hexagon has vertices at a distance aa from the origin, and the outer hexagon has vertices at a distance 2a2a from the origin. Two vertices of each hexagon lie on the y-axis.

1. Initial Electric Field:

For a regular polygon with identical charges at its vertices, the electric field at its center is zero due to symmetry. Since we have two such concentric hexagons, the net electric field at the origin due to all 12 charges initially present would be zero. Let E⃗initial\vec{E}_{\text{initial}} be the electric field at the origin due to all 12 charges.

E⃗initial=0⃗\vec{E}_{\text{initial}} = \vec{0}

2. Charges Removed:

Two charges are removed:

  • A charge from the inner hexagon at position (0,−a)(0, -a). Let's call this charge q1q_1.
  • A charge from the outer hexagon at position (0,2a)(0, 2a). Let's call this charge q2q_2.

3. Electric Field due to Remaining Charges:

Let E⃗final\vec{E}_{\text{final}} be the electric field at the origin after removing these two charges.

The principle of superposition states that the total electric field is the vector sum of the fields due to individual charges.

If E⃗q1\vec{E}_{q_1} is the field produced by charge q1q_1 at the origin, and E⃗q2\vec{E}_{q_2} is the field produced by charge q2q_2 at the origin, then:

E⃗final=E⃗initial−E⃗q1−E⃗q2\vec{E}_{\text{final}} = \vec{E}_{\text{initial}} - \vec{E}_{q_1} - \vec{E}_{q_2}

Since E⃗initial=0⃗\vec{E}_{\text{initial}} = \vec{0}, we have:

E⃗final=−(E⃗q1+E⃗q2)\vec{E}_{\text{final}} = - (\vec{E}_{q_1} + \vec{E}_{q_2})

4. Calculate E⃗q1\vec{E}_{q_1}:

Charge q1=+qq_1 = +q is at (0,−a)(0, -a). The distance from the origin is r1=ar_1 = a.

The electric field at the origin due to this charge points from the charge towards the origin.

The vector from (0,−a)(0, -a) to (0,0)(0, 0) is (0,0)−(0,−a)=(0,a)=aj^(0, 0) - (0, -a) = (0, a) = a\hat{j}.

So, the unit vector is j^\hat{j}.

E⃗q1=kq1r12j^=k(+q)a2j^=kqa2j^\vec{E}_{q_1} = \frac{k q_1}{r_1^2} \hat{j} = \frac{k(+q)}{a^2} \hat{j} = \frac{kq}{a^2} \hat{j}

5. Calculate E⃗q2\vec{E}_{q_2}:

Charge q2=+qq_2 = +q is at (0,2a)(0, 2a). The distance from the origin is r2=2ar_2 = 2a.

The electric field at the origin due to this charge points from the charge towards the origin.

The vector from (0,2a)(0, 2a) to (0,0)(0, 0) is (0,0)−(0,2a)=(0,−2a)=−2aj^(0, 0) - (0, 2a) = (0, -2a) = -2a\hat{j}.

So, the unit vector is −j^-\hat{j}.

E⃗q2=kq2r22(−j^)=k(+q)(2a)2(−j^)=kq4a2(−j^)=−kq4a2j^\vec{E}_{q_2} = \frac{k q_2}{r_2^2} (-\hat{j}) = \frac{k(+q)}{(2a)^2} (-\hat{j}) = \frac{kq}{4a^2} (-\hat{j}) = -\frac{kq}{4a^2} \hat{j}

6. Calculate E⃗final\vec{E}_{\text{final}}:

Substitute the values of E⃗q1\vec{E}_{q_1} and E⃗q2\vec{E}_{q_2} into the equation for E⃗final\vec{E}_{\text{final}}:

E⃗final=−(kqa2j^−kq4a2j^)\vec{E}_{\text{final}} = - \left( \frac{kq}{a^2} \hat{j} - \frac{kq}{4a^2} \hat{j} \right)

E⃗final=−kq(1a2−14a2)j^\vec{E}_{\text{final}} = - kq \left( \frac{1}{a^2} - \frac{1}{4a^2} \right) \hat{j}

To combine the terms in the parenthesis, find a common denominator:

E⃗final=−kq(44a2−14a2)j^\vec{E}_{\text{final}} = - kq \left( \frac{4}{4a^2} - \frac{1}{4a^2} \right) \hat{j}

E⃗final=−kq(4−14a2)j^\vec{E}_{\text{final}} = - kq \left( \frac{4-1}{4a^2} \right) \hat{j}

E⃗final=−3kq4a2j^\vec{E}_{\text{final}} = - \frac{3kq}{4a^2} \hat{j}

7. Magnitude of the Electric Field:

The magnitude of the electric field is:

∣E⃗final∣=3kq4a2|\vec{E}_{\text{final}}| = \frac{3kq}{4a^2}

Substitute the value of Coulomb's constant k=14πϵ0k = \frac{1}{4\pi\epsilon_0}:

∣E⃗final∣=3q4a2(14πϵ0)|\vec{E}_{\text{final}}| = \frac{3q}{4a^2} \left( \frac{1}{4\pi\epsilon_0} \right)

∣E⃗final∣=3q16πϵ0a2|\vec{E}_{\text{final}}| = \frac{3q}{16\pi\epsilon_0a^2}