Answer:
Explanation:
The voltage of a disconnected charged capacitor increases when the plate area is decreased.
When plate area decreases , capacitance C decreases , but charge Q remains constant .
Q = C V where C is capacitance and V is voltage .
when C decreases , V increases for keeping Q constant .
So the statement is true.
The electric field is dependent on the charge density on the plates.
This statement is true .
The voltage of a connected charged capacitor remains the same when the plate area is decreased .
For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .
So the statement is true .
Answer:
The maximum safe depth in salt water is 3758.2 m.
Explanation:
Given that,
Diameter = 20 cm
Radius = 10 cm
Thickness = 9.0 cm
Force 
Inside pressure = 1.0 atm
We need to calculate the area
Using formula of area
Put the value into the formula
We need to calculate the pressure
Using formula of pressure
Put the value into the formula
We need to calculate the maximum depth
Using equation of pressure
Put the value into the formula
Hence, The maximum safe depth in salt water is 3758.2 m.
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
Answer:
a). 53.75 N and 101.92 N
b). 381.44 N and 723.25 N
Explanation:
a).
ρ
, 
, 
b).
ρ, 
, 
