Reasoning: Potential difference (volt) is defined as the work done per unit charge.
$\mathbf{1}\mathbf{\ }\mathbf{V =}$ $\frac{\mathbf{1\ J}}{\mathbf{1\ C}}$
Hence, 1 volt $=$ 1 joule per coulomb.

Reasoning: For two equal positive charges, the electric fields at the midpoint are equal in magnitude but opposite in direction, so they cancel each other. Therefore, the net electric field is zero at the midpoint.

Reasoning: All four charges are at the same distance from the centre, so the magnitude of the electric field due to each charge is proportional to the magnitude of the charge.
Consider the direction of the electric field at the centre:
● At B (+q):
Field is away from the positive charge (towards the bottom-right).
● At C (−q):
Field is towards the negative charge (towards the top-right).
● At A (−2q):
Field is towards the negative charge (towards the bottom-left) with twice the magnitude.
● At D (+2q):
Field is away from the positive charge (towards the top-left) with twice the magnitude.
⮚ Adding the vectors:
● The vertical components cancel each other.
● The horizontal components add towards the left.
Hence, the resultant electric field (and therefore the force on a positive charge placed at the centre) is horizontal towards the left, which is perpendicular to side $\mathbf{AB}$.

Reasoning: The bob experiences:
Weight $\mathbf{= mg}$ (downward) ; Electric force $\mathbf{= qE}$ (horizontal) Since these forces are perpendicular, the tension is their resultant:
$\mathbf{T =}$ ${\mathbf{\lbrack}\mathbf{(mg)}^{\mathbf{2}}\mathbf{+}\mathbf{(gE)}^{\mathbf{2}}\mathbf{\rbrack}}^{\frac{\mathbf{1}}{\mathbf{2}}}$

Reasoning: The electric field points from South to North. A negative charge experiences a force opposite to the electric field. Therefore, it moves towards the South.

Reasoning: Electric field lines are always perpendicular to equipotential surfaces and never lie along them. Therefore, no electric field line connects two points that are at the same potential.

Reasoning: Coulomb’s law describes the force between stationary electric charges, which is the subject of electrostatics.

Reasoning: According to Coulomb’s law: $\mathbf{F =}$ $\frac{\mathbf{k\ }\mathbf{q}_{\mathbf{1}}\mathbf{q}_{\mathbf{2}}}{\mathbf{r}^{\mathbf{2}}}$
Initially: $\mathbf{F =}$ $\frac{\mathbf{k}\mathbf{q}^{\mathbf{2}}}{\mathbf{r}^{\mathbf{2}}}$ After changing: Charges become $\mathbf{2q}$ each. Distance becomes $\mathbf{2r}$.
$\mathbf{F}^{\mathbf{'}}\mathbf{=}$ $\frac{\mathbf{k(2q)(2q)}}{\mathbf{(2r}\mathbf{)}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{4k}\mathbf{q}^{\mathbf{2}}}{\mathbf{4}\mathbf{r}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{k}\mathbf{q}^{\mathbf{2}}}{\mathbf{r}^{\mathbf{2}}}\mathbf{= F}$
Hence, the force remains unchanged.