Electric Field and Potential: Problem 1.14, 1.15 & 1.16 with detailed Solutions

1.14) An electron of mass m₁, initially at rest, covers a certain distance in a uniform electric field in time t₁. A proton of mass m₂, also, takes time t₂ to move through an equal distance (starting from rest) in this uniform electric field. Neglecting the effect of gravity, the ratio t₂/t₁ nearly equals
    A) 1
    B) (m₂/m₁)⁰·⁵
    C) (m₁/m₂)⁰·⁵
    D) 1836

Let the strength of uniform electric field be E. As we know an electron carries a negative charge -e and a proton has a positive charge +e (where e represents the quantum of charge, e = 1.6 × 10⁻¹⁹ C, approx.).

As the electric field is uniform, the charged particles will experience constant electric forces. This implies, the charged particles will undergo uniformly accelerated motion. So the basic equations of motion are applicable here.

|Acceleration| of the electron, a₁ = eE/m₁ and that of the proton, a₂ = eE/m₂. Let d₁ be the distance travelled by the electron in time t₁ and d₂ be the distance traveled by the proton in time t₂. Then,

1.15) An alpha particle having energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach will be of the order
    A) 1 Angstrom
    B) 10⁻¹⁰ cm
    C) 10⁻¹² cm
    D) 10⁻¹⁵ cm

At the instant of closest approach, the velocity of alpha particle and hence the kinetic energy is zero. This means, the whole of its initial kinetic energy is converted into electric potential energy (at the closest approach). Therefore, on applying the principle of conservation of energy, we can write

On further simplifying this expression we get r = 5.3 × 10⁻¹² cm.

1.16) Two point charges q₁ and q₂ are located 3 m apart, and their sum of charges is 10 μC. If force of attraction between them is found to be 0.075 N then the values of q₁ and q₂ respectively, are :
    A) 5 μC, 5 μC
    B) 15 μC, -5 μC
    C) 5 μC, 15 μC
    D) -15 μC, 5 μC

Stated in the problem is that the sum of the two charges is 10μC. From this statement only, the options C and D are eliminated (sum of charges comes out to be 20μC in option C and -10μC in option D).

The remaining options A and B hold that statement (sum of the charges equals 10μC for both the options). We have to choose the correct one from these two options A and B.

It is also said in the question that the force of attraction between them is 0.075N. Thus the two point charges are attracting each other which implies that the charges must be of opposite nature (one positive and the other negative). Because only unlike charges attract each other. Hence, option A cannot be the correct answer as both the charges are of same nature (both positive).

Now, we are only left with the option B which must be the correct choice. You may further verify that the magnitude of Coulomb force between the point charges 15μC and -5μC (the option B) comes out to be 0.075N.

Hindi Hearts

Previous Post Next Post