Answer:
k = 6.72
Explanation:
K of paper = 3.7
k of air = 1
Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So
V = Q / C = 2.5
Capacity becomes C / 3.7 in air .
capacity becomes C/3.7 when paper is replaced by air .
V₁ = Q / (C/3.7)
= 3.7 Q/C
3.7 x 2.5
= 9.25 V
In the second case ,
capacitance due to new unknown dielectric k
= C/3.7 x k
= kC / 3.7 ( Capacitance in air is C/3.7 )
V ( new ) = Q / ( kC/3.7 )
= 3.7 Q/kC
.55 x 2.5 = 3.7 x( 2.5 / k )
k = 3.7 / .55
= 6.72
I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 = 4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s
Therefore; 4 V1 = 0.5 × 3
4V1= 1.5
V1= 1.5/4
= 0.375 m/s
Answer:
Explanation:
Here two charges are placed at distance "d" apart
now the net value of electric field at some position between two charges will be ZERO
so we will have
electric field due to charge 1 = electric field due to charge 2
Let the position where net field is zero will lie at distance "r" from q1
now we will have
now square root both sides
now we have
so we have
