Answer:
c) quadruple in magnitude
Explanation:
The power dissipated in the circuit is given by:
where
I is the current in the circuit
R is the total resistance of the circuit
In this problem:
- The current is doubled: I' = 2 I
- The resistance is kept constant: R' = R
So, the power dissipated is
so, the power dissipated increase by a factor 4 (quadruples).
Answer:
A) i) E =α/ [2πrL(εo)]
ii) E=0
iii) E = α/(πrεo)
The graph between E and r for the 3 cases is attached to this answer ;
B) i) charge on the inner surface per unit length = - α
ii) charge per unit length on the outer surface = 2α
Explanation:
A) i) For r < a, the charge is in the cavity and takes a shape of the cylinder. Thus, applying gauss law;
EA = Q(cavity) / εo
Now, Qcavity = αL
So, E(2πrL) = αL/εo
Making E the subject of the formula, we have;
E =α/ [2πrL(εo)]
ii) For a < r < b; since the distance will be in the bulk of the conductor, therefore, inside the conductor, the electric field will be zero.
So, E=0
iii) For r > b; the total enclosed charge in the system is the difference between the net charge and the charge in the inner surface of the cylinder.
Thus, Qencl = Qnet - Qinner
Qinner will be the negative of Qnet because it should be in the opposite charge of the cavity in order for the electric field to be zero. Thus;
Qencl = αL - (-αL) = 2αL
Thus, applying gauss law;
EA = Qencl / εo
Thus, E = Qencl / Aεo
E = 2αL/Aεo
Since A = 2πrL,
E = 2αL/2πrLεo = α/(πrεo)
B) i) The charge on the cavity wall must be the opposite of the point charge. Therefore, the charge per unit length in the inner surface of the tube will be - α
ii)Net charge per length for tube is +α and there is a charge of - α on the inner surface. Thus charge per unit length on the outer surface will be = +α - (- α) = 2α
