(a) 1.07J
(b) 3.09J
<h2>Explanation:</h2>In an electric field, the change in potential energy (ΔU) of a particle results in a corresponding reversed change in kinetic energy (ΔK). i.e
ΔU = -ΔK
But;
ΔU = U₂ - U₁
ΔK = K₂ - K₁
Equation (i) can be re-written as;
U₂ - U₁ = - ( K₂ - K₁)
U₂ - U₁ = K₁ - K₂ --------------------(ii)
Where;
U₁ = potential energy between a particle of charge q and a point charge Q, at some initial position r₁
U₂ = potential energy between a particle of charge q and a point charge Q, at some later position r₂
K₁ = kinetic energy between a particle of charge q and a point charge Q, at some initial position r₁
K₂ = kinetic energy between a particle of charge q and a point charge Q, at some later position r₂
The potential energy U between a particle of charge q and a point charge Q, at some point, r, is given as;
U = k x q x Q / r ---------------(iii)
Where;
k = constant = 8.99 x 10⁹ Nm²/C²
Using equation (iii) we can write U₁ and U₂ as;
U₁ = k x q x Q / r₁
U₂ = k x q x Q / r₂
Also,
K₁ = 0 [since the particle starts from rest]
K₂ = unknown = kinetic energy at the instant the particle has moved 34cm
Substitute these values of U₁, U₂, K₁ and K₂ into equation (i) as follows;
( k x q x Q / r₂ ) - ( k x q x Q / r₁ ) = 0 - K₂
( k x q x Q / r₂ ) - ( k x q x Q / r₁ ) = - K₂
k x q x Q [ -
] = - K₂ -------------------(iv)
From the question;
r₁ = 70cm = 0.7m
q = +8.2μC = +8.2 x 10⁻⁶ C
(a) When Q is +31μC = +31 x 10⁻⁶ C
Since the two charges are positive, the particle will move 34cm further away (repulsion) i.e from 70cm to 104cm. Therefore;
r₂ = 70 + 34 = 104cm = 1.04m
Substitute the value of Q, q, r₁, r₂ and k into equation (iv) as follows;
8.99 x 10⁹ x 8.2 x 10⁻⁶ x 31 x 10⁻⁶ [ -
] = -K₂
8.99 x 10⁹ x 8.2 x 10⁻⁶ x 31 x 10⁻⁶ [0.96 - 1.43] = -K₂
8.99 x 10⁹ x 8.2 x 10⁻⁶ x 31 x 10⁻⁶ [-0.47] = -K₂
8.99 x 10⁹ x 8.2 x 10⁻⁶ x 31 x 10⁻⁶ x [-0.47] = -K₂
-1.07 = -K₂
K₂ = 1.07J
Therefore, the kinetic energy at that instant is 1.07 J
(b) When Q is -31μC = -31 x 10⁻⁶ C
Since the two charges have different polarities (one is positive and the other is negative), the particle will move 34cm closer (attraction) i.e from 70cm to 36cm. Therefore;
r₂ = 70 - 34 = 36cm = 0.36m
Substitute the value of Q, q, r₁, r₂ and k into equation (iv) as follows;
8.99 x 10⁹ x 8.2 x 10⁻⁶ x -31 x 10⁻⁶ [ -
] = -K₂
8.99 x 10⁹ x 8.2 x 10⁻⁶ x -31 x 10⁻⁶ [2.78 - 1.43] = -K₂
8.99 x 10⁹ x 8.2 x 10⁻⁶ x -31 x 10⁻⁶ [1.35] = -K₂
8.99 x 10⁹ x 8.2 x 10⁻⁶ x -31 x 10⁻⁶ x 1.35 = -K₂
-3.09 = -K₂
K₂ = 3.09 J
Therefore, the kinetic energy at that instant is 3.09 J
Answer:
Explanation:
According to the question
net force F = 2.20×10^6 N
displacement
from figure , the horizontal forces are same in magnitude and opposite direction.
so , neglect these two forces.
we can take only vertical components of the force.
total force F' = F cos 19° + F cos 19°
= 2×F×cos 19° ................. (1
therefore , total work is
W = F'S
= (2F cos19)×S
