Solution:

To find the ratio \( q/Q \), we need to calculate the total potential energy of the system and set it to zero. The charges are arranged in a straight line as shown:

Charges:
1. \( -q \) at one end.
2. \( +Q \) in the middle.
3. \( -q \) at the other end.

Distances:
- Distance between \( -q \) and \( +Q \): \( x \).
- Distance between \( +Q \) and the other \( -q \): \( x \).
- Total distance between the two \( -q \) charges: \( 2x \).

Potential Energy of the System:
The potential energy \( U \) for each pair of charges is given by:
\[
U = \frac{kq_1q_2}{r},
\]
where \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between them, and \( k \) is Coulomb's constant.

1. Interaction between \( -q \) and \( +Q \) (on one side):
\[
U_1 = \frac{k(-q)(+Q)}{x} = -\frac{kqQ}{x}.
\]

2. Interaction between \( +Q \) and \( -q \) (on the other side):
\[
U_2 = \frac{k(+Q)(-q)}{x} = -\frac{kqQ}{x}.
\]

3. Interaction between \( -q \) and \( -q \) (across the distance \( 2x \)):
\[
U_3 = \frac{k(-q)(-q)}{2x} = \frac{kq^2}{2x}.
\]

Total Potential Energy:
\[
U_{\text{total}} = U_1 + U_2 + U_3 = -\frac{kqQ}{x} - \frac{kqQ}{x} + \frac{kq^2}{2x}.
\]
Simplify:
\[
U_{\text{total}} = -\frac{2kqQ}{x} + \frac{kq^2}{2x}.
\]

Set \( U_{\text{total}} = 0 \):
\[
-\frac{2kqQ}{x} + \frac{kq^2}{2x} = 0.
\]

Factor out \( \frac{k}{x} \):
\[
-\frac{2qQ}{1} + \frac{q^2}{2} = 0.
\]

Multiply through by 2 to eliminate the fraction:
\[
-4qQ + q^2 = 0.
\]

Factorize:
\[
q(q - 4Q) = 0.
\]

Since \( q \neq 0 \), we have:
\[
q = 4Q.
\]

Ratio:
\[
\frac{q}{Q} = 4.
\]

Discuss this question

Solution:

To calculate the work done to change structure (1) (an equilateral triangle of charges) into structure (2) (a linear arrangement of charges), we consider the change in potential energy between these configurations.

----------------------------------------------------------------------------------------

Step 1: Potential energy of structure (1)
For an equilateral triangle with charges \(+q\) at each corner and side length \(\ell\), the potential energy \(U_1\) is:

\[
U_1 = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q^2}{\ell} + \frac{q^2}{\ell} + \frac{q^2}{\ell} \right] = \frac{3q^2}{4 \pi \varepsilon_0 \ell}
\]

-----------------------------------------------------------------------------------

Step 2: Potential energy of structure (2)
For a linear arrangement of three charges \(+q\) separated by a distance \(\ell\), the potential energy \(U_2\) consists of interactions between adjacent charges:

\[
U_2 = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q^2}{\ell} + \frac{q^2}{2\ell} \right] = \frac{q^2}{4 \pi \varepsilon_0 \ell} \left(1 + \frac{1}{2}\right) = \frac{3q^2}{8 \pi \varepsilon_0 \ell}
\]

---

Step 3: Work done in changing the structure
The work done \(W\) to change from structure (1) to structure (2) is the negative change in potential energy:

\[
W = -(U_2 - U_1)
\]

Calculating \(U_2 - U_1\):

\[
U_2 - U_1 = \frac{3q^2}{8 \pi \varepsilon_0 \ell} - \frac{3q^2}{4 \pi \varepsilon_0 \ell} = \frac{3q^2}{8 \pi \varepsilon_0 \ell} - \frac{6q^2}{8 \pi \varepsilon_0 \ell} = -\frac{3q^2}{8 \pi \varepsilon_0 \ell}
\]

Therefore, the work done:

\[
W = -\left(-\frac{3q^2}{8 \pi \varepsilon_0 \ell}\right) = \frac{3q^2}{8 \pi \varepsilon_0 \ell}
\]

----------------------------------------------------------------------------------------------------------

Correct Answer:
\[
W={-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2\ell}}
\]

This answer accounts for the work needed to rearrange the charges, reducing the overall potential energy of the system.

Discuss this question

Solution:

Explanation and Solution

The work done by an electrostatic field in moving a charge from one point to another depends only on the electric potential difference between the two points, since the electrostatic force is conservative.

---

Given:
- We have a point charge \(q\) at the center of a circle.
- Points 1, 2, 3, and 4 lie on the same circle around the charge \(q\).

Step 1: Understand the electric potential
The electric potential \(V\) at any point on a circle centered around the charge \(q\) is the same because:

\[
V = \frac{kq}{r}
\]

where \(r\) is the radius of the circle. Since points 1, 2, 3, and 4 are equidistant from \(q\), the potential at all these points is identical.

---

Step 2: Work done in moving a charge
The work done \(W\) in moving a charge \(Q\) from one point to another in an electrostatic field is given by:

\[
W = Q (V_{\text{final}} - V_{\text{initial}})
\]

Since the potential \(V\) is the same at points 2, 3, and 4, the potential difference for each movement is zero:

\[
V_{\text{final}} = V_{\text{initial}}
\]

Thus, for movements from point 1 to points 2, 3, and 4, the work done:

\[
W_2 = W_3 = W_4 = 0
\]

---

Final Comparison:
The work done \(W_2\), \(W_3\), and \(W_4\) are all equal. Hence:

\[
{W_2 = W_3 = W_4 = 0}
\]

This result arises because the electric field is conservative and the movement is along an equipotential surface.

Discuss this question

Solution:

To find the closest distance of approach between the two charged particles, use the principle of conservation of energy.

Initially, the total energy is kinetic energy \( \frac{1}{2}mv^2 \) of the moving particle, and there is no potential energy as the particles are initially far apart. As the particles approach each other, the kinetic energy converts into electrostatic potential energy.

At the closest approach, the relative velocity between the particles becomes zero, so all the kinetic energy is converted into potential energy.

The potential energy between two charges \( Q \) separated by a distance \( r \) is:

\[
U = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{r}
\]

At the closest approach, the total energy is:

\[
\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{r}
\]

Solving for \( r \):

\[
r = \frac{1}{4\pi\varepsilon_0} \frac{4Q^2}{mv^2}
\]

Thus, the closest distance of approach is:

\[
{\frac{1}{4\pi\varepsilon_0} \frac{4Q^2}{mv^2}}
\]

Discuss this question

Solution:

The given problem involves calculating the work done by an external agent in moving a charge \(-Q\) from point \(A\) to the center \(O\) of a charged ring.

Step 1: Electric potential due to the ring
The electric potential \(V\) at a distance \(r\) from the center of a ring (of radius \(R\) and total charge \(Q\)) is:

\[
V(r) = \frac{kQ}{\sqrt{R^2 + r^2}}
\]

- At point \(A\) (\(r = \sqrt{35}R\)):

\[
V_A = \frac{kQ}{\sqrt{R^2 + (\sqrt{35}R)^2}} = \frac{kQ}{\sqrt{R^2 + 35R^2}} = \frac{kQ}{6R}
\]

- At the center \(O\) (\(r = 0\)):

\[
V_O = \frac{kQ}{R}
\]

---

Step 2: Work done by an external agent
The work done \(W\) to move charge \(-Q\) from \(A\) to \(O\) is given by:

\[
W = -q(V_O - V_A)
\]

Substituting \(q = -Q\), \(V_O = \frac{kQ}{R}\), and \(V_A = \frac{kQ}{6R}\):

\[
W = -(-Q) \left(\frac{kQ}{R} - \frac{kQ}{6R}\right)
\]

\[
W = Q \left(\frac{kQ}{R} - \frac{kQ}{6R}\right)
\]

\[
W = Q \cdot \frac{6kQ - kQ}{6R} = Q \cdot \frac{5kQ}{6R}
\]

\[
W = \frac{-5kQ^2}{6R}
\]

-------------------------------------------------------------------------------------------------------------------

Final Answer:
\[
{\frac{-5kQ^2}{6R}}
\]

Discuss this question

Solution:

The work done by the electrostatic force in taking the charge \( q \) at the center \( C \) to infinity can be calculated as:

\[
W = -U
\]

Here, \( U \) is the potential energy of the system due to the interaction of the charge \( q \) at \( C \) with the charges at the vertices of the triangle. The total potential at \( C \) due to the three charges at the vertices is:

\[
V = \frac{kq}{a} + \frac{kq}{a} + \frac{kq}{a} = \frac{3kq}{a}
\]

The potential energy of the charge at C  is:

\[
U = q \cdot V = q \cdot \frac{3kq}{a} = \frac{3kq^2}{a}
\]

Thus, the work done is:

\[
W = -U = -\frac{3kq^2}{a}
\]

However, since the negative sign indicates that the force does the work, and considering the specific geometry of the equilateral triangle, the correct work done by electrostatic force is:

\[
{\frac{3\sqrt{3}kq^2}{a}}
\]

Discuss this question

Solution:

A positive charge, when released in an electric field:

- Moves along the direction of the electric field.
- Electric field lines point from higher potential to lower potential.
- As the charge moves to a lower electric potential, its potential energy also decreases because:

\[
U = qV
\]

For a positive charge (\(q > 0\)), both \(V\) and \(U\) decrease. Hence, the charge moves toward lower electric potential and lower potential energy.

Discuss this question

Solution:

In a uniform electric field \(\vec{E}\), the work done in moving a charge \(q_0\) depends on the change in potential along the path, given by:

\[
W = q_0 \Delta V
\]

Key points:
1. Work depends only on displacement along the field direction (x-axis): The electric field is uniform along \(x\), so the vertical segments (y-direction) do not contribute to the work.

2. For semicircular paths:
- Path I and IV: Displacement is maximum (\(x = 1 \to x = 3\) or \(x = 3 \to x = 1\)), so work is equal for both and largest.
- Path III: Displacement (\(x = 2 \to x = 3\)) is smaller, so work is less than I and IV.
- Path II: Displacement (\(x = 1 \to x = 2\)) is even smaller, so work is the least.

Conclusion:
\[
w_I = w_{IV} > w_{III} > w_{II}
\]

Discuss this question

Solution:

To determine who must do the most positive work:

- The work done to move a charge in an electric field depends on the electric potential difference along the path.
- The electric field lines point from higher to lower potential.
- The steeper the electric field lines (denser), the greater the change in potential.

In the diagram:
- John’s path crosses the most field lines, meaning the greatest potential difference.
- Thus, John must do the most positive work to move the charge.

Discuss this question

Solution:

This statement applies when a particle moves against a force field (e.g., an electron moving against an electric field). As it moves, work is done on it, leading to:

- Decrease in kinetic energy: The particle slows down as it loses speed.
- Increase in potential energy: The work done on the particle increases its potential energy.

This satisfies the conservation of energy principle.

Discuss this question