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Twenty-seven drops of same size are charged at 220 220 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
medium
Electrostatic Potential and Capacitance
2021
physics
Explanation To solve this problem, we need to find the potential of the bigger drop formed by the combination of 27 smaller drops. \\
Given: \\
โข Each small drop is charged at 220 220 220 V. \\
โข Total number of small drops n = 27. n = 27. \\
n = 27. Let's denote: \\
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V s = 220 V_s = 220 V s โ = 220 V as the potential of each small drop.
โข
as the potential of the bigger drop.
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as the radius of each small drop.
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as the radius of the bigger drop.
The volume of a single small drop is
4 3 ฯ r s 3 . \frac{4}{3} \pi r_s^3. \\
3 4 โ ฯ r s 3 โ . The total volume of 27 small drops is
27 ร 4 3 ฯ r s 3 . 27 \times \frac{4}{3} \pi r_s^3. \\
27 ร 3 4 โ ฯ r s 3 โ . When these drops combine to form a bigger drop, the volume of the bigger drop is:
4 3 ฯ r b 3 = 27 ร 4 3 ฯ r s 3 . \\
\frac{4}{3} \pi r_b^3 = 27 \times \frac{4}{3} \pi r_s^3. \\
3 4 โ ฯ r b 3 โ = 27 ร 3 4 โ ฯ r s 3 โ . Simplifying, we find:
r b 3 = 27 r s 3 . \\
r_b^3 = 27 r_s^3. \\
r b 3 โ = 27 r s 3 โ . Taking the cube root of both sides:
r b = 3 r s . \\
r_b = 3 r_s. \\
r b โ = 3 r s โ . The potential of a charged sphere is given by
V = k Q r , V = \frac{kQ}{r}, V = r k Q โ , where
is the charge and
is the radius.
Since the potential of each small drop is
V s = k Q s r s = 220 V_s = \frac{kQ_s}{r_s} = 220 V s โ = r s โ k Q s โ โ = 220 V,
the charge on each small drop is
Q s = 220 r s k . Q_s = \frac{220 r_s}{k}. \\
Q s โ = k 220 r s โ โ . The total charge on the bigger drop is
Q b = 27 Q s . Q_b = 27 Q_s. \\
Q b โ = 27 Q s โ . Substituting for
Q s : Q b = 27 ร 220 r s k . Q_s: \\
Q_b = 27 \times \frac{220 r_s}{k}. \\
Q s โ : Q b โ = 27 ร k 220 r s โ โ . The potential of the bigger drop is
V b = k Q b r b . V_b = \frac{kQ_b}{r_b}. \\
V b โ = r b โ k Q b โ โ . Substituting for
and
r b : V b = k ร 27 ร 220 r s k 3 r s . r_b: \\
V_b = \frac{k \times 27 \times \frac{220 r_s}{k}}{3 r_s}. \\
r b โ : V b โ = 3 r s โ k ร 27 ร k 220 r s โ โ โ . Simplifying:
V b = 27 ร 220 r s 3 r s . V b = 9 ร 220. V b = 1980 \\
V_b = \frac{27 \times 220 r_s}{3 r_s}. \\
V_b = 9 \times 220. \\
V_b = 1980 V b โ = 3 r s โ 27 ร 220 r s โ โ . V b โ = 9 ร 220. V b โ = 1980 V.
Therefore, the potential of the bigger drop is
V, which corresponds to Option 4.