Here, as the charge is uniformly distributed in the sphere, we will consider s as an area of the sphere which is, s=4πr2 and r is radius of the gaussian surface shown in the figure above. From this, it can be seen that
Element. element cannot be simplerfide
Answer:
0.0006091222 m
Explanation:
q = Charge = 42 pC
V = Voltage = 620 V
= Permittivity of free space = 
Electric potential is given by (at r = R)
The radius of the drop is 0.0006091222 m
The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J
Assuming air friction is negligible,
a = - 9.8 m / s²
u = 31.4 m / s
s = 30 m
v² = u² + 2 a s
v² = 31.4² + ( 2 * - 9.8 * 30 )
v² = 985.96 - 588
v² = 397.96 m / s
KE = 1 / 2 m v²
KE = 1 / 2 * 0.155 * 397.96
KE = 0.0775 * 397.96
KE = 30.85 J
Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J
To know more about kinetic energy
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