Question

Object X and object Y, of charge +Q and −2Q, respectively, are separated by a distance r0, as shown in the figure.

(a) Determine an equation for the magnitude and direction of the electric force that is exerted on object Y from object X in terms of Q, r0, and physical constants, as appropriate.

(b) The charge of object X is changed so that it has a charge of −4Q, and the separation distance between the two objects is changed to 2r0. In a clear, coherent paragraph-length response that may also contain figures and/or equations, predict how the electric force exerted on object Y from object X changes from part (a), and predict how the net charge of the two object system changes from part (a).

Answer 1

Answer:

(a) The magnitude of the force electric force F = k ×(+Q) × (-2·Q)/r₀² = -2·Q²/(r₀²) which can be written as follows;

$F = \dfrac{k \times (+ Q) \times (-2\cdot Q)}{r_0^2} = -\dfrac{k \times 2 \times Q^2}{r_0^2}$

Given that the charge of Y is twice the charge on X, we have the charge X will move towards the charge Y which is the +x direction

(b) The force, F, acting between the charges is given as follows;

$F = \dfrac{k \times Q_1 \times Q_2}{r_0^2}$

When, Q₁ = X = -4Q, Q₂ = Y = -2Q, and r₀ = 2·r₀ we have;

$F = \dfrac{k \times (-4\cdot Q) \times (-2\cdot Q)}{(2 \cdot r_0)^2} = \dfrac{k \times 8 \times Q^2}{4 \cdot r_0^2} = \dfrac{k \times 2 \times Q^2}{r_0^2}$

Therefore, the electric force exerted on object Y by the object X is the same and acts in an opposite direction to the force of X on Y in (a)

The net charge of the two objects in part (a) is Q - 2Q = -Q

The net charge of the two objects in part (b) is -4Q - 2Q = -6Q

Therefore, the net charge increases by a factor of 6 in part (b)

Explanation: