201. A charge \(q\) experiences two electrostatic forces of equal magnitude \(F\) with an angle \(60^\circ\) between them. What is the magnitude of the resultant force?
ⓐ. \(F\)
ⓑ. \(\sqrt{2}F\)
ⓒ. \(\sqrt{3}F\)
ⓓ. \(2F\)
Correct Answer: \(\sqrt{3}F\)
Explanation: \( \textbf{Given forces:} \) Two forces have equal magnitude \(F\).
\( \textbf{Angle between them:} \) \(\theta=60^\circ\).
\( \textbf{Resultant formula:} \) \(R=\sqrt{F^2+F^2+2F^2\cos\theta}\).
\( \textbf{Substitute angle:} \) \(R=\sqrt{F^2+F^2+2F^2\cos60^\circ}\).
\( \textbf{Use value:} \) \(\cos60^\circ=\frac{1}{2}\).
\( \textbf{Simplification:} \) \(R=\sqrt{2F^2+F^2}=\sqrt{3F^2}\).
\( \textbf{Final result:} \) \(R=\sqrt{3}F\).
202. Three equal positive charges are placed at the vertices of an equilateral triangle. What is the direction of the net force on one charge due to the other two?
ⓐ. Along the outward angle bisector
ⓑ. Along the side joining the other two charges
ⓒ. Toward the centre of the triangle
ⓓ. Zero because all three charges are equal
Correct Answer: Along the outward angle bisector
Explanation: Consider one positive charge at a vertex of the equilateral triangle. The other two positive charges repel it along the two sides meeting at that vertex. Since the triangle is equilateral, the two forces have equal magnitudes. These equal forces are symmetrically placed about the angle bisector at the chosen vertex. Their components perpendicular to the angle bisector cancel. Their components along the angle bisector add. Because both interactions are repulsive, the resultant points away from the opposite side of the triangle. Therefore, the net force is along the angle bisector directed outward.
203. A charge is in electrostatic equilibrium under the action of several other charges. Which condition must be satisfied?
ⓐ. The charge must be positive.
ⓑ. The algebraic sum of all source charges must be zero.
ⓒ. The net electrostatic force on the charge must be zero.
ⓓ. The charge must be placed exactly midway between two charges.
Correct Answer: The net electrostatic force on the charge must be zero.
Explanation: Electrostatic equilibrium of a charge means that the resultant electrostatic force on it is zero. Individual forces may still act on the charge, but their vector sum must cancel. The condition is written as \(\vec{F}_{\text{net}}=0\). The charge need not be positive; a negative charge can also be in equilibrium if the forces balance. The algebraic sum of source charges need not be zero because force depends on magnitudes, signs, distances, and directions. A midpoint is an equilibrium point only in special symmetric cases. Therefore, the necessary condition is zero net force.
204. Two fixed like charges \(+9Q\) and \(+Q\) are separated by distance \(d\). A small positive test charge is to be placed between them where the net force is zero. What is the ratio of its distance from \(+9Q\) to its distance from \(+Q\)?
ⓐ. \(1:3\)
ⓑ. \(3:1\)
ⓒ. \(1:9\)
ⓓ. \(9:1\)
Correct Answer: \(3:1\)
Explanation: \( \textbf{Let distances be:} \) Distance from \(+9Q\) is \(x\), and distance from \(+Q\) is \(y\).
\( \textbf{Force balance condition:} \) The force due to \(+9Q\) must equal the force due to \(+Q\) in magnitude.
\( \textbf{Coulomb equality:} \) \(k\frac{9Qq}{x^2}=k\frac{Qq}{y^2}\).
\( \textbf{Cancel common factors:} \) \(\frac{9}{x^2}=\frac{1}{y^2}\).
\( \textbf{Cross relation:} \) \(x^2=9y^2\).
\( \textbf{Taking positive square root:} \) \(x=3y\).
\( \textbf{Ratio:} \) \(x:y=3:1\).
\( \textbf{Final result:} \) The point is three times farther from \(+9Q\) than from \(+Q\), so the ratio is \(3:1\).
205. A small charge is placed at a point where two electrostatic forces on it are equal in magnitude and opposite in direction. Which statement is valid?
ⓐ. The charge is in equilibrium at that point.
ⓑ. The two source charges must have equal signs.
ⓒ. The force on each source charge is zero.
ⓓ. The electric charges must all be neutral.
Correct Answer: The charge is in equilibrium at that point.
Explanation: If two forces on the same charge are equal in magnitude and opposite in direction, their vector sum is zero. A zero net force is the condition for equilibrium of that charge. This statement concerns the charge placed at that point, not necessarily the source charges. The source charges may still experience forces due to other charges. Equal and opposite forces on one charge should not be confused with action-reaction forces acting on different bodies. The source charges do not have to be neutral. Therefore, the charge is in equilibrium at that point.
206. Two fixed like charges \(+4Q\) and \(+9Q\) are separated by \(10\,\text{cm}\). At what distance from \(+4Q\) should a small positive test charge be placed between them so that the net force on it is zero?
ⓐ. \(2\,\text{cm}\)
ⓑ. \(4\,\text{cm}\)
ⓒ. \(5\,\text{cm}\)
ⓓ. \(6\,\text{cm}\)
Correct Answer: \(4\,\text{cm}\)
Explanation: \( \textbf{Given arrangement:} \) The charges \(+4Q\) and \(+9Q\) are like charges separated by \(10\,\text{cm}\).
\( \textbf{Location of equilibrium:} \) For like charges, the zero-force point for a test charge lies between them.
\( \textbf{Let distance from \(+4Q\):} \) \(x\).
\( \textbf{Distance from \(+9Q\):} \) \(10-x\), in \(\text{cm}\).
\( \textbf{Force-balance condition:} \) \(k\frac{4Qq}{x^2}=k\frac{9Qq}{(10-x)^2}\).
\( \textbf{Cancel common factors:} \) \(\frac{4}{x^2}=\frac{9}{(10-x)^2}\).
\( \textbf{Square-root step:} \) \(\frac{2}{x}=\frac{3}{10-x}\).
\( \textbf{Solving:} \) \(2(10-x)=3x\), so \(20-2x=3x\), giving \(x=4\,\text{cm}\).
\( \textbf{Final result:} \) The test charge should be placed \(4\,\text{cm}\) from \(+4Q\).
207. Two fixed charges \(+Q\) and \(+16Q\) are separated by distance \(d\). A zero-force point for a small positive test charge lies between them. Which statement about its position is correct?
ⓐ. It is midway between the two charges.
ⓑ. It is closer to the charge \(+16Q\).
ⓒ. It is outside the two charges.
ⓓ. It is closer to the charge \(+Q\).
Correct Answer: It is closer to the charge \(+Q\).
Explanation: Between two like charges, forces on a test charge due to the two fixed charges are in opposite directions. For equilibrium, the stronger source charge must be farther away so that its larger charge is balanced by greater distance. Here \(+16Q\) is much larger than \(+Q\). Therefore, the zero-force point must lie closer to the smaller charge \(+Q\). If the point were midway, the force due to \(+16Q\) would be much larger. If it were closer to \(+16Q\), the imbalance would become even greater. Hence, the equilibrium point lies between the charges but nearer to \(+Q\).
208. Charges \(+Q\) and \(-9Q\) are fixed \(20\,\text{cm}\) apart. A small positive test charge is placed on the line outside the pair on the side of \(+Q\). How far from \(+Q\) is the zero-force point?
ⓐ. \(10\,\text{cm}\)
ⓑ. \(20\,\text{cm}\)
ⓒ. \(30\,\text{cm}\)
ⓓ. \(40\,\text{cm}\)
Correct Answer: \(10\,\text{cm}\)
Explanation: \( \textbf{Given charges:} \) The fixed charges are \(+Q\) and \(-9Q\), separated by \(20\,\text{cm}\).
\( \textbf{Location logic:} \) For unequal unlike charges, the zero-force point lies outside on the side of the smaller magnitude charge.
\( \textbf{Let distance from \(+Q\):} \) \(x\), outside on the side away from \(-9Q\).
\( \textbf{Distance from \(-9Q\):} \) \(x+20\), in \(\text{cm}\).
\( \textbf{Force-balance condition:} \) \(k\frac{Qq}{x^2}=k\frac{9Qq}{(x+20)^2}\).
\( \textbf{Cancel common factors:} \) \(\frac{1}{x^2}=\frac{9}{(x+20)^2}\).
\( \textbf{Taking square root:} \) \(\frac{1}{x}=\frac{3}{x+20}\).
\( \textbf{Solving:} \) \(x+20=3x\), so \(2x=20\), giving \(x=10\,\text{cm}\).
\( \textbf{Final result:} \) The zero-force point is \(10\,\text{cm}\) from \(+Q\) on the outside.
209. Which statement about electrostatic equilibrium of a charge in a simple two-charge arrangement is most accurate?
ⓐ. It always occurs at the midpoint of any two charges.
ⓑ. It requires the two source charges to be equal and opposite.
ⓒ. The vector sum of forces on the charge must be zero.
ⓓ. It occurs only when no individual electrostatic force acts.
Correct Answer: The vector sum of forces on the charge must be zero.
Explanation: Electrostatic equilibrium means that the net electrostatic force on the charge is zero. This does not require each individual force to be zero. Several non-zero forces may act and still cancel by vector addition. The point of equilibrium is the midpoint only in special symmetric cases, such as equal like charges with a test charge at the midpoint. Unequal charges shift the equilibrium point closer to the smaller like charge or outside for unequal unlike charges. Equal and opposite source charges do not automatically give a finite zero-force point between them. Therefore, the essential condition is \(\vec{F}_{\text{net}}=0\).
210. Two equal positive charges are fixed at points \(A\) and \(B\). A small positive charge is placed exactly midway between them. If it is displaced slightly toward \(A\), what is the qualitative nature of its equilibrium along the line \(AB\)?
ⓐ. Unstable; a nearer charge repels it away from the midpoint.
ⓑ. Stable, because the nearer charge pulls it back toward the midpoint.
ⓒ. Neutral, because both forces remain equal after displacement.
ⓓ. Stable, because positive charges always return to the midpoint.
Correct Answer: Stable, because the nearer charge pulls it back toward the midpoint.
Explanation: A small displacement toward A makes repulsion from A stronger, so the charge is pushed back toward the midpoint; equilibrium is stable along AB.
211. Which definition correctly gives the electric field at a point?
ⓐ. \(\vec{E}=q_0\vec{F}\), where \(q_0\) is any source charge
ⓑ. \(\vec{E}=\frac{q_0}{\vec{F}}\), where \(q_0\) is a negative test charge
ⓒ. \(\vec{E}=\frac{\vec{F}}{q}\), where \(q\) is the source charge creating the field
ⓓ. \(\vec{E}=\frac{\vec{F}}{q_0}\), where \(q_0\) is a small positive test charge
Correct Answer: \(\vec{E}=\frac{\vec{F}}{q_0}\), where \(q_0\) is a small positive test charge
Explanation: \( \textbf{Definition:} \) Electric field at a point is force per unit positive test charge placed at that point.
\( \textbf{Formula:} \) \(\vec{E}=\frac{\vec{F}}{q_0}\).
\( \textbf{Meaning of \(q_0\):} \) The test charge \(q_0\) is taken small and positive.
\( \textbf{Why small:} \) A small test charge reduces disturbance of the source charge distribution.
\( \textbf{Why positive:} \) The direction of \(\vec{E}\) is defined as the direction of force on a positive test charge.
\( \textbf{Source distinction:} \) The field is created by source charges, but \(q_0\) is used only to measure the field.
\( \textbf{Final result:} \) The correct definition is \(\vec{E}=\frac{\vec{F}}{q_0}\).
212. A positive test charge \(q_0=2.0\times10^{-6}\,\text{C}\) experiences a force \(6.0\times10^{-3}\,\text{N}\) at a point. What is the magnitude of the electric field at that point?
ⓐ. \(1.2\times10^{-8}\,\text{N C}^{-1}\)
ⓑ. \(3.0\times10^3\,\text{N C}^{-1}\)
ⓒ. \(1.2\times10^{-2}\,\text{N C}^{-1}\)
ⓓ. \(3.0\times10^{-3}\,\text{N C}^{-1}\)
Correct Answer: \(3.0\times10^3\,\text{N C}^{-1}\)
Explanation: \( \textbf{Given force:} \) \(F=6.0\times10^{-3}\,\text{N}\).
\( \textbf{Given test charge:} \) \(q_0=2.0\times10^{-6}\,\text{C}\).
\( \textbf{Required quantity:} \) Electric field magnitude \(E\).
\( \textbf{Definition used:} \) \(E=\frac{F}{q_0}\).
\( \textbf{Substitution:} \) \(E=\frac{6.0\times10^{-3}}{2.0\times10^{-6}}\,\text{N C}^{-1}\).
\( \textbf{Coefficient simplification:} \) \(\frac{6.0}{2.0}=3.0\).
\( \textbf{Power simplification:} \) \(\frac{10^{-3}}{10^{-6}}=10^3\).
\( \textbf{Final result:} \) \(E=3.0\times10^3\,\text{N C}^{-1}\).
213. Why is a test charge used in defining electric field taken to be small?
ⓐ. To make the electric field zero at every point
ⓑ. To make the test charge negative automatically
ⓒ. To avoid significant disturbance of the source charges
ⓓ. To change the unit of field from \(\text{N C}^{-1}\) to \(\text{C N}^{-1}\)
Correct Answer: To avoid significant disturbance of the source charges
Explanation: Electric field at a point is meant to describe the influence of the source charges at that point. A test charge is introduced only to detect the force that would act there. If the test charge is large, it may exert significant forces on the source charges and disturb their arrangement. Then the measured force may not represent the original field of the source charges alone. Taking \(q_0\) small reduces this disturbance. The test charge is also conventionally positive so that field direction is defined clearly. Its smallness is not meant to make the field zero. Therefore, the test charge is taken small to avoid altering the field being measured.
214. Which statement correctly compares electric field and electrostatic force?
ⓐ. Field is due to source charges; force also depends on the placed charge.
ⓑ. Electric field and force are always identical quantities with the same unit.
ⓒ. Electric field exists only if the test charge is negative.
ⓓ. Force is independent of the charge placed in the field.
Correct Answer: Field is due to source charges; force also depends on the placed charge.
Explanation: Electric field at a point is a property of the source charge arrangement. It describes the force that would be experienced by a unit positive test charge placed there. The actual force on a charge placed in the field is given by \(\vec{F}=q\vec{E}\). Therefore, the force depends on both the electric field and the value of the charge placed there. A larger charge experiences a larger force in the same field. A negative charge experiences force opposite to the field direction. Electric field has unit \(\text{N C}^{-1}\), while force has unit \(\text{N}\). Thus, field and force are related but not identical.
215. A charge \(q=+3.0\,\mu\text{C}\) is placed at a point where the electric field is \(2.0\times10^4\,\text{N C}^{-1}\) toward east. What is the force on the charge?
ⓐ. \(6.0\times10^{-2}\,\text{N}\) toward west
ⓑ. \(6.0\times10^{-2}\,\text{N}\) toward east
ⓒ. \(1.5\times10^{-10}\,\text{N}\) toward east
ⓓ. \(1.5\times10^{-10}\,\text{N}\) toward west
Correct Answer: \(6.0\times10^{-2}\,\text{N}\) toward east
Explanation: \( \textbf{Given charge:} \) \(q=+3.0\,\mu\text{C}=3.0\times10^{-6}\,\text{C}\).
\( \textbf{Given field:} \) \(E=2.0\times10^4\,\text{N C}^{-1}\), toward east.
\( \textbf{Required:} \) Force on the charge.
\( \textbf{Relation:} \) \(\vec{F}=q\vec{E}\).
\( \textbf{Magnitude:} \) \(F=|q|E=(3.0\times10^{-6})(2.0\times10^4)\,\text{N}\).
\( \textbf{Simplification:} \) \(F=6.0\times10^{-2}\,\text{N}\).
\( \textbf{Direction check:} \) Since \(q\) is positive, \(\vec{F}\) is in the same direction as \(\vec{E}\).
\( \textbf{Final result:} \) The force is \(6.0\times10^{-2}\,\text{N}\) toward east.
216. A charge \(q=-4.0\,\mu\text{C}\) is placed in a uniform electric field of magnitude \(5.0\times10^3\,\text{N C}^{-1}\) directed upward. What is the force on the charge?
ⓐ. \(2.0\times10^{-2}\,\text{N}\) down
ⓑ. \(2.0\times10^{-2}\,\text{N}\) up
ⓒ. \(1.25\times10^9\,\text{N}\) down
ⓓ. \(1.25\times10^9\,\text{N}\) up
Correct Answer: \(2.0\times10^{-2}\,\text{N}\) down
Explanation: \( \textbf{Given charge:} \) \(q=-4.0\,\mu\text{C}=-4.0\times10^{-6}\,\text{C}\).
\( \textbf{Electric field:} \) \(E=5.0\times10^3\,\text{N C}^{-1}\), directed upward.
\( \textbf{Force relation:} \) \(\vec{F}=q\vec{E}\).
\( \textbf{Magnitude calculation:} \) \(F=|q|E=(4.0\times10^{-6})(5.0\times10^3)\,\text{N}\).
\( \textbf{Simplification:} \) \(F=20\times10^{-3}\,\text{N}=2.0\times10^{-2}\,\text{N}\).
\( \textbf{Direction rule:} \) A negative charge experiences force opposite to the electric field direction.
\( \textbf{Field direction:} \) The field is upward, so the force is downward.
\( \textbf{Final result:} \) The force is \(2.0\times10^{-2}\,\text{N}\) downward.
217. What is the SI unit of electric field?
ⓐ. \(\text{C N}^{-1}\)
ⓑ. \(\text{N m}^2\text{C}^{-2}\)
ⓒ. \(\text{N C}^{-1}\)
ⓓ. \(\text{C m}^{-2}\)
Correct Answer: \(\text{N C}^{-1}\)
Explanation: \( \textbf{Definition of field:} \) Electric field is force per unit positive test charge.
\( \textbf{Formula:} \) \(E=\frac{F}{q_0}\).
\( \textbf{Unit of force:} \) Force is measured in \(\text{N}\).
\( \textbf{Unit of charge:} \) Charge is measured in \(\text{C}\).
\( \textbf{Unit of electric field:} \) \([E]=\frac{\text{N}}{\text{C}}=\text{N C}^{-1}\).
\( \textbf{Distinction:} \) \(\text{N m}^2\text{C}^{-2}\) is the unit of Coulomb's constant \(k\), not electric field.
\( \textbf{Final result:} \) The SI unit of electric field is \(\text{N C}^{-1}\).
218. What is the direction of electric field at a point near a positive point charge?
ⓐ. Toward the positive charge
ⓑ. Away from the positive charge
ⓒ. Perpendicular to the radius from the charge
ⓓ. Opposite to the force on a positive test charge
Correct Answer: Away from the positive charge
Explanation: The direction of electric field is defined as the direction of force on a positive test charge. A positive source charge repels a positive test charge. Therefore, a positive test charge placed near a positive source charge would be pushed away from the source charge. The electric field direction must follow that force direction. Hence, the field due to a positive point charge is radially outward. The field is not directed toward the positive charge. It is also not perpendicular to the radial line in the point-charge case. Therefore, the field is directed away from the positive charge.
219. What is the direction of electric field at a point near a negative point charge?
ⓐ. Toward the negative charge
ⓑ. Away from the negative charge
ⓒ. Along the velocity of the negative charge
ⓓ. Opposite to the force on a positive test charge
Correct Answer: Toward the negative charge
Explanation: Electric field direction is defined using a positive test charge. A negative source charge attracts a positive test charge. Therefore, the force on the positive test charge is directed toward the negative source charge. Since electric field points in the direction of force on a positive test charge, the field is toward the negative charge. This is why field lines terminate on negative charges. The field direction is not decided by the velocity of the source charge in electrostatics. Thus, the field near a negative point charge is radially inward.
220. A negative charge is placed in an electric field directed toward the east. What is the direction of force on the negative charge?
ⓐ. Toward east
ⓑ. Toward north
ⓒ. Zero because the charge is negative
ⓓ. Toward west
Correct Answer: Toward west
Explanation: Electric field direction is the direction of force on a positive test charge. A positive charge placed in an eastward electric field would experience force toward east. A negative charge behaves oppositely because \(\vec{F}=q\vec{E}\), and \(q<0\) reverses the direction of the force relative to \(\vec{E}\). Therefore, the force on the negative charge is opposite to the eastward field. The direction opposite to east is west. The force is not zero merely because the charge is negative. Hence, the force on the negative charge is toward west.
