Class 12 Physics MCQs | Chapter 2: Electrostatic Potential and Capacitance – Part 2

Explanation: Electric potential due to a point charge is a scalar and is written as \(V=\frac{kq}{r}\). Since \(q\) may be positive or negative, the potential may also be positive or negative. Electric field, however, is a vector quantity. When only its magnitude is written, the expression is \(E=\frac{k|q|}{r^2}\), which is non-negative. The direction of the field must be stated separately: away from a positive charge and toward a negative charge. Using \(q^2\) would give the wrong dependence and lose the physical sign information for potential. The sign of scalar potential is not the same as vector direction. This distinction prevents mixing up \(V\) and \(\vec{E}\).

Explanation: \(\textbf{Given:}\) \(V=-120\,\text{V}\) and \(r=0.50\,\text{m}\). \(\textbf{Required:}\) Electric field magnitude \(E\). \(\textbf{Point-charge relations:}\) \(V=\frac{kq}{r}\) and \(E=\frac{k|q|}{r^2}\). \(\textbf{Combining relations:}\) Since \(|V|=\frac{k|q|}{r}\), the field magnitude is \(E=\frac{|V|}{r}\). \(\textbf{Substitution:}\) \(E=\frac{120\,\text{V}}{0.50\,\text{m}}\). \(\textbf{Calculation:}\) \(E=240\,\text{V m}^{-1}\). \(\textbf{Unit equivalence:}\) \(1\,\text{V m}^{-1}=1\,\text{N C}^{-1}\). \(\textbf{Final result:}\) The electric field magnitude is \(240\,\text{N C}^{-1}\).

Explanation: The electric potential due to a point charge depends only on distance from the charge, not on direction around it. Its expression is \(V=\frac{kq}{r}\). If two points are at the same distance \(r\) from the same point charge, they have the same potential. Being on the opposite side changes the direction of the electric field vector, but it does not change the scalar potential. Potential has no direction to reverse. The sign of the potential is set by the sign of the source charge, not by the side of the charge on which the point lies. Therefore, the second point also has potential \(V\).

Explanation: \(\textbf{Given:}\) \(V_1=+400\,\text{V}\), \(r_1=0.10\,\text{m}\), and \(V_2=+100\,\text{V}\). \(\textbf{Required:}\) New distance \(r_2\). \(\textbf{Potential-distance relation:}\) For a fixed point charge, \(V\propto\frac{1}{r}\). \(\textbf{Ratio form:}\) \(V_1r_1=V_2r_2\). \(\textbf{Solving for distance:}\) \(r_2=\frac{V_1r_1}{V_2}\). \(\textbf{Substitution:}\) \(r_2=\frac{(400)(0.10)}{100}\,\text{m}\). \(\textbf{Calculation:}\) \(r_2=0.40\,\text{m}\). \(\textbf{Reasonableness check:}\) The potential becomes one-fourth, so the distance should become four times larger. \(\textbf{Final result:}\) The distance is \(0.40\,\text{m}\).

Explanation: \(\textbf{Given contributions:}\) \(V_1=+30\,\text{V}\) and \(V_2=+50\,\text{V}\). \(\textbf{Principle:}\) Electric potential obeys superposition as an algebraic scalar sum. \(\textbf{No direction needed:}\) Potential is not resolved into components. \(\textbf{Net potential:}\) \(V=V_1+V_2\). \(\textbf{Substitution:}\) \(V=30\,\text{V}+50\,\text{V}\). \(\textbf{Calculation:}\) \(V=80\,\text{V}\). \(\textbf{Final result:}\) The net potential at \(P\) is \(+80\,\text{V}\).

Explanation: \(\textbf{Potential contributions:}\) \(V_1=+45\,\text{V}\), \(V_2=-20\,\text{V}\), and \(V_3=-10\,\text{V}\). \(\textbf{Superposition rule:}\) Net potential is the algebraic sum of individual potentials. \(\textbf{Expression:}\) \(V=V_1+V_2+V_3\). \(\textbf{Substitution:}\) \(V=45-20-10\,\text{V}\). \(\textbf{Simplification:}\) \(45-20=25\), and \(25-10=15\). \(\textbf{Sign check:}\) The positive contribution is larger than the total negative contribution. \(\textbf{Final result:}\) The resultant potential is \(+15\,\text{V}\).

Explanation: \(\textbf{Potential due to one charge:}\) A point charge \(q\) at distance \(r\) gives \(V=\frac{kq}{r}\). \(\textbf{Many-charge rule:}\) Electric potential is scalar, so individual potentials add algebraically. \(\textbf{Sign treatment:}\) The signs of \(q_1\), \(q_2\), and \(q_3\) must be retained. \(\textbf{Total potential:}\) The sum is \(V=V_1+V_2+V_3\). \(\textbf{Substitution:}\) \(V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}+k\frac{q_3}{r_3}\). \(\textbf{Common factor:}\) \(V=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}\right)\). \(\textbf{Final result:}\) The required expression is \(V=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}\right)\).

Explanation: \(\textbf{Given charges:}\) \(q_1=+3.0\,\mu\text{C}=3.0\times10^{-6}\,\text{C}\) and \(q_2=-1.0\,\mu\text{C}=-1.0\times10^{-6}\,\text{C}\). \(\textbf{Distance:}\) Both charges are at \(r=0.60\,\text{m}\) from \(P\). \(\textbf{Formula:}\) \(V=k\left(\frac{q_1}{r}+\frac{q_2}{r}\right)=\frac{k(q_1+q_2)}{r}\). \(\textbf{Net algebraic charge in numerator:}\) \(q_1+q_2=+2.0\times10^{-6}\,\text{C}\). \(\textbf{Substitution:}\) \(V=\frac{(9.0\times10^9)(2.0\times10^{-6})}{0.60}\,\text{V}\). \(\textbf{Calculation:}\) \((9.0\times10^9)(2.0\times10^{-6})=18.0\times10^3\). \(\textbf{Division:}\) \(V=\frac{18.0\times10^3}{0.60}=3.0\times10^4\,\text{V}\). \(\textbf{Final result:}\) The potential at \(P\) is \(+3.0\times10^4\,\text{V}\).

Explanation: \(\textbf{Charge arrangement:}\) The charges are \(+q\) and \(-q\). \(\textbf{Distance condition:}\) Both are at the same distance \(r\) from point \(P\). \(\textbf{Potential from \(+q\):}\) \(V_+=\frac{kq}{r}\). \(\textbf{Potential from \(-q\):}\) \(V_-=-\frac{kq}{r}\). \(\textbf{Algebraic addition:}\) \(V=V_++V_-=\frac{kq}{r}-\frac{kq}{r}\). \(\textbf{Result:}\) \(V=0\). \(\textbf{Important caution:}\) The electric field at \(P\) may or may not be zero, depending on directions of field contributions. \(\textbf{Final result:}\) The net electric potential at \(P\) is \(0\).

Explanation: \(\textbf{Known contribution:}\) The first potential contribution is \(+120\,\text{V}\). \(\textbf{Required total:}\) The net potential should be \(0\). \(\textbf{Algebraic addition:}\) Let the second contribution be \(V_2\). \(\textbf{Equation:}\) \(+120\,\text{V}+V_2=0\). \(\textbf{Solving:}\) \(V_2=-120\,\text{V}\). \(\textbf{Meaning:}\) A negative contribution of equal magnitude cancels the positive scalar potential. \(\textbf{Final result:}\) The other contribution must be \(-120\,\text{V}\).

Explanation: \(\textbf{Charges:}\) The charges are \(+2q\), \(-q\), and \(-q\). \(\textbf{Distance condition:}\) Each charge is at the same distance \(r\) from \(O\). \(\textbf{Potential formula:}\) Each contribution is \(V_i=\frac{kq_i}{r}\). \(\textbf{Total potential:}\) \(V=\frac{k}{r}(2q-q-q)\). \(\textbf{Algebraic sum of charges:}\) \(2q-q-q=0\). \(\textbf{Result:}\) \(V=\frac{k}{r}(0)=0\). \(\textbf{Reasoning check:}\) Equal distance allows direct cancellation through the algebraic sum of charges. \(\textbf{Final result:}\) The potential at \(O\) is \(0\).

Explanation: \(\textbf{Given:}\) \(q_1=+2.0\,\mu\text{C}=2.0\times10^{-6}\,\text{C}\), \(r_1=0.20\,\text{m}\), \(q_2=-4.0\,\mu\text{C}=-4.0\times10^{-6}\,\text{C}\), and \(r_2=0.40\,\text{m}\). \(\textbf{Required:}\) Net potential at \(P\). \(\textbf{Superposition formula:}\) \(V=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)\). \(\textbf{First ratio:}\) \(\frac{q_1}{r_1}=\frac{2.0\times10^{-6}}{0.20}=10\times10^{-6}\,\text{C m}^{-1}\). \(\textbf{Second ratio:}\) \(\frac{q_2}{r_2}=\frac{-4.0\times10^{-6}}{0.40}=-10\times10^{-6}\,\text{C m}^{-1}\). \(\textbf{Algebraic sum:}\) \(10\times10^{-6}-10\times10^{-6}=0\). \(\textbf{Potential:}\) \(V=k(0)=0\). \(\textbf{Final result:}\) The net potential at \(P\) is \(0\).

Explanation: \(\textbf{Charge arrangement:}\) Three identical positive charges are placed symmetrically around \(O\). \(\textbf{Distance condition:}\) Each charge is at the same distance \(r\) from \(O\). \(\textbf{Potential due to one charge:}\) \(V_1=\frac{kq}{r}\). \(\textbf{Scalar addition:}\) Electric potential is a scalar, so the three contributions add algebraically. \(\textbf{Total potential:}\) \(V=V_1+V_2+V_3\). \(\textbf{Substitution:}\) \(V=\frac{kq}{r}+\frac{kq}{r}+\frac{kq}{r}\). \(\textbf{Simplification:}\) \(V=\frac{3kq}{r}\). \(\textbf{Final result:}\) The electric potential at \(O\) is \(\frac{3kq}{r}\).

Explanation: \(\textbf{Geometry:}\) The midpoint is at distance \(a\) from each charge because the separation is \(2a\). \(\textbf{Potential from first charge:}\) \(V_1=\frac{kq}{a}\). \(\textbf{Potential from second charge:}\) \(V_2=\frac{kq}{a}\). \(\textbf{Addition rule:}\) Electric potential is scalar, so the two positive contributions add directly. \(\textbf{Total potential:}\) \(V=V_1+V_2\). \(\textbf{Substitution:}\) \(V=\frac{kq}{a}+\frac{kq}{a}\). \(\textbf{Simplification:}\) \(V=\frac{2kq}{a}\). \(\textbf{Final result:}\) The electric potential at the midpoint is \(\frac{2kq}{a}\).

Explanation: At the midpoint, the distances from \(+q\) and \(-q\) are equal. The potential due to \(+q\) is \(+\frac{kq}{a}\), while the potential due to \(-q\) is \(-\frac{kq}{a}\). These scalar contributions cancel, so the net electric potential is zero. Electric field is different because it is a vector. At the midpoint, the field due to \(+q\) points away from \(+q\), and the field due to \(-q\) points toward \(-q\). These two field directions are the same along the line joining the charges, so the fields add instead of canceling. Hence the midpoint has zero potential but non-zero electric field.

Explanation: \(\textbf{Configuration:}\) Four equal positive charges are placed symmetrically at the corners of a square. \(\textbf{Distance from centre:}\) Each charge is at the same distance \(r\) from the centre. \(\textbf{Potential due to one charge:}\) \(V_1=\frac{kq}{r}\). \(\textbf{Superposition:}\) Potential is scalar, so all four contributions add algebraically. \(\textbf{Total potential:}\) \(V=4\left(\frac{kq}{r}\right)\). \(\textbf{Simplification:}\) \(V=\frac{4kq}{r}\). \(\textbf{Field contrast:}\) The electric field at the centre may cancel by symmetry, but potential does not cancel because all contributions are positive. \(\textbf{Final result:}\) The potential at the centre is \(\frac{4kq}{r}\).

Explanation: \(\textbf{Distance condition:}\) All four charges are at the same distance \(r\) from \(O\). \(\textbf{Potential contribution:}\) Each charge contributes \(V_i=\frac{kq_i}{r}\). \(\textbf{Total potential:}\) \(V=\frac{k}{r}(q+q-q-q)\). \(\textbf{Algebraic charge sum:}\) \(q+q-q-q=0\). \(\textbf{Potential value:}\) \(V=\frac{k}{r}(0)=0\). \(\textbf{Important idea:}\) This cancellation happens because potential is a scalar and signs are added algebraically. \(\textbf{Final result:}\) The net electric potential at \(O\) is \(0\).

Explanation: \(\textbf{Charge positions:}\) The point lies between \(+4Q\) and \(-Q\). \(\textbf{Distances:}\) Distance from \(+4Q\) is \(x\), so distance from \(-Q\) is \(d-x\). \(\textbf{Zero-potential condition:}\) The algebraic sum of potentials must be zero. \(\textbf{Equation:}\) \(\frac{k(4Q)}{x}+\frac{k(-Q)}{d-x}=0\). \(\textbf{Cancel common factors:}\) \(kQ\) cancels, giving \(\frac{4}{x}-\frac{1}{d-x}=0\). \(\textbf{Rearrangement:}\) \(\frac{4}{x}=\frac{1}{d-x}\). \(\textbf{Cross multiplication:}\) \(4(d-x)=x\). \(\textbf{Solving:}\) \(4d-4x=x\), so \(5x=4d\) and \(x=\frac{4d}{5}\). \(\textbf{Reasonableness check:}\) The zero-potential point between unequal opposite charges lies closer to the smaller-magnitude charge \(-Q\), so it must be farther from the larger charge \(+4Q\). \(\textbf{Final result:}\) \(x=\frac{4d}{5}\).

Explanation: \(\textbf{Position of zero potential:}\) The point is between the two charges. \(\textbf{Distances:}\) Distance from \(+Q\) is \(x\), and distance from \(-3Q\) is \(d-x\). \(\textbf{Potential equation:}\) \(\frac{kQ}{x}+\frac{k(-3Q)}{d-x}=0\) \(\textbf{Cancel common factors:}\) \(kQ\) cancels from the equation. \(\textbf{Reduced equation:}\) \(\frac{1}{x}-\frac{3}{d-x}=0\). \(\textbf{Rearrangement:}\) \(\frac{1}{x}=\frac{3}{d-x}\). \(\textbf{Cross multiplication:}\) \(d-x=3x\). \(\textbf{Solving:}\) \(d=4x\), so \(x=\frac{d}{4}\). \(\textbf{Final result:}\) The zero-potential point between the charges is at \(\frac{d}{4}\) from \(+Q\).

Explanation: \(\textbf{Charges:}\) The charges are \(+2Q\) and \(+Q\). \(\textbf{Equal distance:}\) Both are at distance \(r\) from \(P\). \(\textbf{Potential from \(+2Q\):}\) \(V_1=\frac{k(2Q)}{r}=\frac{2kQ}{r}\). \(\textbf{Potential from \(+Q\):}\) \(V_2=\frac{kQ}{r}\). \(\textbf{Scalar addition:}\) Both contributions are positive, so they add. \(\textbf{Total potential:}\) \(V=\frac{2kQ}{r}+\frac{kQ}{r}\). \(\textbf{Simplification:}\) \(V=\frac{3kQ}{r}\). \(\textbf{Final result:}\) The potential is positive and equal to \(\frac{3kQ}{r}\).