Answer:
v = 1.15*10^{7} m/s
Explanation:
given data:
charge/ unit area
plate seperation = 1.69*10^{-2} m
we know that
electric field btwn the plates is
force acting on charge is F = q E
Work done by charge q id
this work done is converted into kinectic enerrgy
solving for v
v = 1.15*10^{7} m/s
Answer:
v = 10 V and E = 2 10³ N/C
Explanation:
The electrical potentials and the electric field at one point are related by the expression
ΔV = - ∫ E. dS
Where the bold indicates vector quantities, E is the electric field and S is the line of displacement of the load, in general displacement is perpendicular to the equipotential lines, which reduces the product scales to the ordinary product.
If the potential difference is the most usual that is V = 10 V, the electric field is
s = 0.5 cm = 0.5 10⁻² m
E = ΔV / S
E = 10/0.5 10⁻²
E = 2 10³ N / C
The membership rose among the baptist and methodists.
1) The coefficient of friction of the surface
2) The weight of that objects
