Complete Question
The complete question is shown on the first uploaded image
Answer:
a it is always zero
b 0
c 
Explanation:ss
Here the net charge is on the outer surface of the conductor thus this means that the net charge inside the conductor is zero
Generally the charge density of a conductor is dependent on the charge per unit area which implies that the charge density is dependent on the net charge so this means that the charge density inside the conductor is zero
Generally the direction of electric field this from the positive charge to the negative charge so from the question we can deduce that the negative charge is located on the surface of the conductor
So We can mathematically define the charge density on the surface of the electric field as
∮
Where E is the electric field
change in unit area
is the negative charge
is the permittivity of free space
So



Where
is the charge density
Answer:
The battery can supply 130 W for 11.75 h
Explanation:
In order to discover the time in wich the battery can supply this energy we need to find how much current is being drawn from it, we do that by using the equation for real power that is P = V*I, since we have V and P we can solve for I as seen bellow:
I = P/V = 130/12 = 10.834 A
We can use this value to find how many hours the power can supply said current. We do that by dividing the current capacity of the battery by the current drawn:
t = 141/12 = 11.75 h
gravitational force of planet exerted on its object near the surface is known as weight
so here we know that gravitational force of mars is much less than the gravitational force of Earth
So on the surface of mars the Weight of objects must be much less than the weight of object on surface of Earth
so here correct answer must be
<em>D. Your weight would decrease.</em>
Answer:
t = 5.56 ms
Explanation:
Given:-
- The current carried in, Iin = 1.000002 C
- The current carried out, Iout = 1.00000 C
- The radius of sphere, r = 10 cm
Find:-
How long would it take for the sphere to increase in potential by 1000 V?
Solution:-
- The net charge held by the isolated conducting sphere after (t) seconds would be:
qnet = (Iin - Iout)*t
qnet = t*(1.000002 - 1.00000) = 0.000002*t
- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:
V = k*qnet / r
Where, k = 8.99*10^9 ..... Coulomb's constant
qnet = V*r / k
t = 1000*0.1 / (8.99*10^9 * 0.000002)
t = 5.56 ms
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