Answer:
The total charge Q of the sphere is 
.
Explanation:
Given that,
Radius = 5 cm
Charge density 
We need to calculate the total charge Q of the sphere
Using formula of charge
Where, 
= charge density
V = volume
Put the value into the formula
Put the value into the formula
Hence, The total charge Q of the sphere is .
Answer:
Explanation:
When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.
When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other .So both the shells lose their charges .The positive half shell also loses all its charges
When we separate the half shells , there will be no deflection in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.
Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
The charge of the copper nucleus is 29 times the charge of one proton:
the charge of the electron is
and their separation is
The magnitude of the electrostatic force between them is given by:
where
is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
Answer:
The correct answer should be
A. 20 Joules
Explanation:
I'm taking the K12 Unit Test: Energy - Part 1 right now
