Four identical 2.5 mF capacitors are connected together electrically. What is the greatest possible capacitance of the combinati
1 answer:
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Answer:
See the answers below.
Explanation:
We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F =m*a
where:
F = force [N] (units of newtons)
m = mass = 1000 [kg]
a = acceleration = 3 [m/s²]
And the weight of any body can be calculated by means of the mass product by gravitational acceleration.
Answer:
(a) 51428.59 J/C
(b) 25714.29 J/C
(c) 0 J/C
Explanation:
Parameters given:
Q1 = 2 * 10^-6 C
Q2 = 2 * 10^-6 C
Q3 = 2 * 10^-6 C
Q4 = 2 * 10^-6 C
=> Q1 = Q2 = Q3 = Q4 = Q
Side of the square = 2m
The center of the square is the midpoint of the diagonals, i.e. Using Pythagoras theorem:
BD² = 2² + 2²
BD² = 8
BD = √(8) = 2.8m
OD = 1.4m
(The attached diagram explains better)
Hence, the distance between the center and each point charge, r, is 1.4m.
Electric Potential, V = kQ/r
k = Coulombs constant
(a) If all charges are positive:
V(Total) = V1 + V2 + V3 + V4
V1 = Potential due to Q1
V2 = Potential due to Q2
V3 = Potential due to Q3
V4 =Potential due to Q4
Since Q1 = Q2 = Q3 = Q4 = Q
=> V1 = V2 = V3 = V4
=> V(Total) = 4V1
V = (4 * 9 * 10^9 * 2 * 10^-6)/1.4
V = 51428.59J/C
(b) If 3 charges are positive and 1 is negative:
Since Q1 = Q2 = Q3 = Q
and Q4 = -Q
The total potential becomes:
V(Total) = V1 + V2 + V3 - V4
Since V1, V2, V3 and V4 have the same value,
V(Total) = V1 + V2
V(Total) = 2V1
V(Total) = (2 * 9 * 10^9 * 2 * 10^-6)/1.4
V(Total) = 25174.29 J/C
(c) Two charges are positive and two are negative:
Since Q1 = Q2 = Q
and Q3 = Q4 = -Q
The total potential becomes:
V(Total) = V1 + V2 - V3 - V4
Since V1, V2, V3 and V4 have the same value,
V(Total) = 0 J/C
