Unit 10: Electric Force, Field, and Potential
Weightage : 15–18%
A gas contains two types of charged particles. Negatively charged particle \(X^{-}\) has mass \(m\) and velocity \(+v_{0}\), as shown in the figure. It collides head-on with positively charged particle \(Y^{+}\) that has mass \(8m\) and velocity \(-v_{0}\). Electrostatic force then holds the particles together.
What is the final velocity of the two-particle system?
Correct Answer: \( \boxed{\mathrm{B}} \)
Since the particles stick together after the collision due to the electrostatic force, the collision is perfectly inelastic. Therefore, the total linear momentum is conserved.
Initial momentum:
\( p_i=mv_{0}+8m(-v_{0})=mv_{0}-8mv_{0}=-7mv_{0}. \)
The combined mass after the collision is
\( m+8m=9m. \)
Applying conservation of momentum,
\( -7mv_{0}=9mv_f. \)
Solving for the final velocity,
\( v_f=-\dfrac{7}{9}v_{0}. \)
The negative sign indicates that the combined particles move in the original direction of the heavier particle \(Y^{+}\).
Therefore, the correct answer is (B).
Three identical, small cork spheres are released from rest in a uniform electric field directed downward toward the floor. Sphere X is uncharged, but spheres Y and Z are charged. Sphere Z remains suspended in the field. Spheres X and Y fall downward, but sphere X takes a longer time to reach the floor.
What are the signs of the charges on spheres Y and Z?
Sphere Y Sphere Z
Correct Answer: \( \boxed{\mathrm{D}} \)
The electric force on a charged object is
\(F_{\mathrm{E}}=qE.\)
Since the electric field is directed downward, a positive charge experiences a downward electric force, while a negative charge experiences an upward electric force.
Sphere Z remains suspended, so the upward electric force must exactly balance its weight:
\(F_{\mathrm{E}}=mg.\)
Therefore, sphere Z must carry a negative charge so that the electric force acts upward.
Sphere X is uncharged, so it falls under gravity alone.
Sphere Y reaches the floor before sphere X, which means it has a greater downward acceleration. This is only possible if the electric force acts downward in addition to gravity.
Therefore, sphere Y must have a positive charge.
Thus,
Sphere Y: Positive
Sphere Z: Negative
Therefore, the correct answer is (D).
As shown above, two particles, each of charge \(+Q\), are fixed at opposite corners of a square that lies in the plane of the page. A positive test charge \(+q\) is placed at a third corner.
If \(F\) is the magnitude of the force on the test charge due to only one of the other charges, what is the magnitude of the net force acting on the test charge due to both of these charges?
Correct Answer: \( \boxed{\mathrm{C}} \)
Each source charge exerts a repulsive force of magnitude \(F\) on the positive test charge.
The two forces are perpendicular to each other: one acts horizontally to the left and the other acts vertically downward.
Using the Pythagorean theorem for vector addition,
\( \displaystyle F_{\mathrm{net}}=\sqrt{F^2+F^2}=\sqrt{2F^2}=F\sqrt{2} \)
Thus, the magnitude of the resultant force is \( \sqrt{2}F \), directed diagonally downward and to the left.
Therefore, the correct answer is (C).
