Learn conservation of linear momentum and mechanical energy, practice closest distance when one charge is fix, and distance of closest approach when both charges are free and fly away.

- Net external force on the system must be zero i.e., \(\Sigma \vec F_{ext}=0 \implies \Delta \vec p=0\)

Here, \(\vec p_i=\) Initial momentum, \(\vec p_f=\) Final momentum

\(\vec p_i=\vec p_f\)

- Charges of the system are free.

Here, \(v_1=\) Velocity of \(q_1\) and \(v_2=\) Velocity of \(q_2\)

- If any of the charge is fixed, momentum conservation is not applied because an external force is required to fix that charge.
- Linear momentum conservation is not applicable.

- Work done by internal non-conservative force and external force is zero.

\(W_{{int}_{nc}}=0\) and \(W_{ext}=0\)

**When all charges are free**

**Reason:** Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

**When one of the charge is fixed**

**Reason:** External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

**Flowchart:**

- Work done by internal non-conservative force and external force is zero.

\(W_{int}=0\) and \(W_{ext}=0\)

**When all charges are free**

**Reason: **Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

**When one of the charge is fixed**

**Reason:** External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

**Flowchart:**

A \(3\times10^8 \,m/sec\)

B \(2\times10^{11}\, m/sec\)

C \(\sqrt3\times10^5\, m/sec\)

D \(1.9\times10^{10}\, m/sec\)

- Work done by internal non-conservative force and external force is zero.

\(W_{int}=0\) and \(W_{ext}=0\)

**When all charges are free**

**Reason:** Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

**When one of the charges is fixed**

**Reason:** External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

**Flowchart :**

A \(\dfrac {Q}{4\pi\epsilon_0v^2 }\)

B \(\dfrac {Qq}{4\pi\epsilon_0v^2}\)

C \(\dfrac {Qq}{2\pi\epsilon_0(mv^2)}\)

D \(\dfrac {Q}{4\pi\epsilon_0m}\)

- Net external force on the system must be zero i.e., \(\Sigma \vec F_{ext}=0 \Rightarrow \Delta \vec p=0\)

Here, \(\vec p_i=\) Initial momentum, \(\vec p_f=\)Final momentum

\(\vec p_i=\vec p_f\)

- Charges of the system are free

Here, \(v_1=\) Velocity of \(q_1\) and \(v_2=\) Velocity of \(q_2\)

- If any of the charge is fixed, momentum conservation is not applied because an external force is required to fix that charge.
- Linear momentum conservation is not applicable.

- Work done by internal non-conservative force and external force is zero.
- \(W_{int}=0\) and \(W_{ext}=0\)

**When all charges are free**

**Reason:** Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

**When one of the charges is fix**

**Reason:** External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

**Flowchart :**

A \(\dfrac{q}{4\pi\epsilon_0a}\)

B \(\dfrac{q^2}{\sqrt {4\pi\epsilon_0ma}}\)

C \(\dfrac{q}{\sqrt {4\pi\epsilon_0a}}\)

D \(q\,\, \sqrt {\dfrac{4+\sqrt2} {8\pi\epsilon_0ma}}\)

- Work done by internal non-conservative force and external force is zero.
- \(W_{int}=0\) and \(W_{ext}=0\)
**When all charges are free**

**Reason: **Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

**When one of the charges is fixed**

**Reason:** External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

**Flowchart :**

A \(\dfrac {q}{\sqrt {2\pi\epsilon_0ma}}\)

B \(\dfrac {q}{\sqrt {4\pi\epsilon_0ma}}\)

C \(\dfrac {q}{\sqrt {\pi\epsilon_0ma}}\)

D \(\dfrac {q}{\sqrt {8\pi\epsilon_0ma}}\)

- Net external force on the system must be zero i.e., \(\Sigma \vec F_{ext}=0 \Rightarrow \Delta \vec p=0\)

here, \(\vec p_i=\) Initial momentum, \(\vec p_f=\) Final momentum

\(\vec p_i=\vec p_f\)

- Charges of the system are free

Here , \(v_1=\) Velocity of \(q_1\) and \(v_2=\) Velocity of \(q_2\)

- If any of the charge is fixed, momentum conservation is not applied because an external force is required to fix that charge.
- Linear momentum conservation is not applicable.

- Work done by internal non-conservative force and external force is zero.

\(W_{int}=0\) and \(W_{ext}=0\)

**When all charges are free**

**Reason:** Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

**When one of the charges is fixed**

**Reason:** External force is present, but work done by this force is zero since charge is fixed, i.e., no displacement occurs. So, mechanical energy conservation is applicable.

**Flowchart:**

A \(\dfrac {q}{ \sqrt {4\pi\epsilon_0am}}\)

B \(q\,\,\sqrt {\dfrac {3} {4\pi\epsilon_0am}}\)

C \(\dfrac {q}{ \sqrt {2\pi\epsilon_0am}}\)

D \(\dfrac {q}{ \sqrt {8\pi\epsilon_0am}}\)

- Work done by internal non-conservative force and external force is zero. \(W_{int}=0\) and \(W_{ext}=0\)
- When all charges are free,

**REASON: **Work done by non-conservative force and external force is zero as they are absent. So, mechanical energy conservation is applicable.

- When one of the charges is fixed,

**REASON:** External force is present, but work done by this force is zero since charge is fixed, i.e. no displacement occurs. So, mechanical energy conservation is applicable.

A \(\dfrac {q^2}{4\pi\epsilon_0mv^2}\)

B \(\dfrac {q}{4\pi\epsilon_0mv^2}\)

C \(\dfrac {q^2}{2\pi\epsilon_0mv^2}\)

D \(\dfrac {q}{2\pi\epsilon_0mv^2}\)

- Consider the potential difference between two points in a uniform electric field be \(\Delta V\), then difference in potential energy is given by \(\Delta U=q\Delta V\)

[Since, potential difference is potential energy per unit charge]

\(\therefore \Delta V=\dfrac {\Delta U}{q}\)

- Let charge q has velocity v
_{1}and v_{2}at point A and B, respectively.

By energy conservation between A and B,

K.E_{A} + U_{A} = K.E_{B} + U_{B}

U_{A }— U_{B} = K.E_{B} – K.E_{A}

\(-\Delta U=\Delta K.E\)

\(-q\Delta V=\Delta K.E\)