The Electric field is zero at a distance 2.492 cm from the origin.
Let A be point where the charge 
C is placed which is the origin.
Let B be the point where the charge 
C is placed. Given that B is at a distance 1 cm from the origin.
Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.
i.e., at distance 'x' from B.
Using Coulomb's law, 
,
= 
k is the Coulomb's law constant.
On substituting the values into the above equation, we get,
Taking square roots on both sides and simplifying and solving for x, we get,
1.67x = 1+x
Therefore, x = 1.492 cm
Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.
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Answer:
6 km is the right answer
hope it helps you
Explanation:
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Answer:
Part 1 
Part 2 
Part 3 
Explanation:
Given
Number of protons 
Radius of nucleus 
Distance of the electrons 
Part 1
Electric field produced by just outside its surface
Part 2
Electric field produced by just outside its surface
Part 3
The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other
hence, the solution is
Part 1
Part 2
Part 3
Answer:
Explanation:
Energy stored in a capacitor
= 1/2 CV²
C is capacitance and V is potential of the capacitor .
When capacitor is charged to 24 V ,
E₁ = 1/2 x 2.4 x 24 x24 = 691.2 J
When it is charged to 12 volt
E₂ = 1/2 CV²
.5 X 2.4 X 12 X12
= 172.8 J
