Electrostatic Potential: Point Charges, Potential Energy & Field Relation: 26 MCQs with answers

B · Volt (J/C)

Potential is work per unit charge, so its unit is the volt, and $1\,\text{V} = 1\,\text{J/C}$. N/C is the unit of electric field, not potential.

B · Negative, because $Q$ is negative

$V = KQ/r$ carries the sign of $Q$. A negative source charge produces negative potential everywhere around it, regardless of the test charge.

B · Zero

Each charge is at distance $x$ from the midpoint. $V_{net} = \frac{KQ}{x} + \frac{K(-Q)}{x} = 0$ because the scalar contributions cancel with signs.

C · $10\,\text{J}$

$U = qV = (2)(5) = 10\,\text{J}$. The potential energy is simply the product of charge and potential.

A · True

Electric potential is a scalar, so contributions from multiple charges add algebraically with signs, unlike the vector addition required for electric field.

B · V = KQ/r

The correct formula is \( V = rac{KQ}{r} \), where \( K \) is a constant, \( Q \) is the charge, and \( r \) is the distance from the charge.

C · Volt

The SI unit of electric potential is the Volt (V), defined as one Joule per Coulomb (1 V = 1 J/C).

B · U = qV

The electric potential energy is given by \( U = qV \), where \( q \) is the charge and \( V \) is the electric potential.

C · The potential is zero at the midpoint

At the midpoint between two equal and opposite charges, the contributions to potential cancel out, resulting in a total potential of zero.

C · The potential is changing

A non-zero electric field indicates that the electric potential is changing, as the field points from high potential to low potential.

C · W = q\Delta V

The relationship is given by \( W = q\Delta V \), where \( W \) is the work done by an external agent in moving charge \( q \).

A · \Delta V = -Ed

In a uniform electric field, the potential difference is given by \( \Delta V = -Ed \), indicating that potential decreases in the direction of the field.

A · No work is done

When moving a charge along an equipotential surface, no work is done because the potential remains constant.

C · The relationship between potential and field

This equation represents the relationship between electric potential and electric field, indicating how potential changes along a path in the field.

D · The potential is constant

When the electric field is zero, the potential can still be defined and is constant in that region, but it does not have to be zero.

C · $\dfrac{2KQ}{x}$

Both charges contribute $+\frac{KQ}{x}$ at the midpoint, so $V_{net} = \frac{2KQ}{x}$. The fields cancel, but the scalar potentials add.

B · It is non-zero and positive

Zero field does not imply zero potential. Both $+Q$ charges contribute positive potential, so $V = \frac{2KQ}{x} \neq 0$ even though $\vec{E} = 0$.

A · $20\,\text{V}$

For a uniform field along the displacement, $|\Delta V| = Ed = (100)(0.2) = 20\,\text{V}$.

A · $12\,\text{J}$

$W_{ext} = q\,\Delta V = (3)(4) = 12\,\text{J}$. The work done by the electric field would be $-12\,\text{J}$, opposite in sign.

D · Toward lower potential (right)

A positive charge spontaneously moves from high to low potential. The field points from high to low, so the force on a positive charge is in that direction.

C · Potential is a scalar, so contributions add algebraically with signs

Because $V$ is a scalar, $V_{net} = K\sum Q_i/r_i$ — no components or angles. Field requires vector addition, which is more involved.

C · It has zero potential energy but feels a non-zero force

At the midpoint $V = 0$, so $U = qV = 0$. But the fields from $+Q$ and $-Q$ point the same way, so $\vec{E} \neq 0$ and the test charge feels a force.

C · The change in potential along the path

By definition, an equipotential surface has the same potential everywhere, so $\Delta V = 0$ along it. The field is perpendicular to the surface and is generally not zero.

D · Field is a vector sum; both fields point the same way and add

The scalar potentials cancel, but the field vectors from $+Q$ and $-Q$ both point from $+$ to $-$ at the midpoint, so they reinforce rather than cancel.

A · Zero, because $\vec{E}\cdot d\vec{r} = 0$

When the path is perpendicular to $\vec{E}$, the dot product $\vec{E}\cdot d\vec{r} = 0$, so the integral vanishes and $\Delta V = 0$ — the path lies on an equipotential.

A · It moves from low to high potential, opposite to a positive charge

The force on a negative charge is opposite to $\vec{E}$, so it accelerates toward higher potential. A positive charge would move the other way.