The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v
<u>Explanation</u>:
Given data,
charge = 5.5 x 10¹² C
k =9.00 x 10⁹
The electric potential V of a point charge can found by,
V= kQ / r
Assuming, r=5.00×10⁻² m
V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m
V= 49.5 x 10⁻³/ 5.00×10⁻²
Electric potential V= 0.099 x 10⁻¹v
Answer:
Part a)
%
Part b)
%
Explanation:
As we know that total power used in the room is given as

here we have






Part a)
Since power supply is at 110 Volt so the current obtained from this supply is given as


now resistance of transmission line



now power loss in line is given as



Now percentage loss is given as


%
Part b)
now same power must have been supplied from the supply station at 110 kV, so we have


now power loss in line is given as



Now percentage loss is given as


%
R = 0.407Ω.
The resistance R of a particular conductor is related to the resistivity ρ of the material by the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of the material.
To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.
We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4. Then:
R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]
R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²
R = 0.407Ω
Answer:
The given statement is false.
Explanation:
For any negative vector

The magnitude of the vector is given by

As we know that square root of any quantity cannot be negative thus we conclude that the right hand term in the above expression cannot be negative hence we conclude that magnitude of any vector cannot be negative.