is electric field or magnetic field responcible For death of a person....if we touch a wire carrying tremendous electric current....?
Answers
Answer:
Answer is at attachment ...
Answer:
Answer:
New force of repulsion = 99 N
Explanation:
Given:
Two point charges repel each other with a force of 100 N
One charge is increased by 10% and the other charge is reduced by 10%
To Find:
The new force of repulsion
Solution:
By Coulomb's law we know that the force between two point charges is given by,
\boxed{\sf F =k\times \frac{q_1q_2}{r^2}}
F=k×
r
2
q
1
q
2
where k is the Coulomb's law constant, F is the force, q₁, q₂ are the point charges and r is the distance between the two charges.
The initial force of repulsion is given by,
\sf F_1=k\times \dfrac{q_1q_2}{r^2} =100\:NF
1
=k×
r
2
q
1
q
2
=100N
Now given that the first charge is increased by 10%
Hence,
(q₁)' = q₁ + q₁ × 10/100
(q₁)' = q₁ + q₁/10
(q₁)' = 11 q₁/10-------(1)
Also by given second charge is decreased by 10%
Hence,
(q₂)' = q₂ - q₂ × 10/100
(q₂)' = 9 q₂/10 ------(2)
Now the new force of repulsion is given by,
\sf F_2=k\times \dfrac{(q_1)'(q_2)'}{r^2}F
2
=k×
r
2
(q
1
)
′
(q
2
)
′
Substitute the values from 1 and 2,
\sf F_2=k\times \dfrac{(11\:q_1/10)(9\:q_2/10)}{r^2}F
2
=k×
r
2
(11q
1
/10)(9q
2
/10)
\sf F_2=\dfrac{99}{100}\times k\times \dfrac{q_1q_2}{r^2}F
2
=
100
99
×k×
r
2
q
1
q
2
Hence,
F₂ = 0.99 × F₁
F₂ = 0.99 × 100
F₂ = 99 N
Hence the new force of repulsion is 99 N.
Explanation:
this answer for your first question