A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
1 answer:
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:
You might be interested in
Answer:
The time taken for the race is 17.20 s.
Explanation:
It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.
Calculate the final speed of the sprinter.
The expression for the equation of the motion is as follows;
Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.
Put u= 0, s=30 m and .
Calculate time taken to cover 30 m distance.
The expression for the equation of motion is as follows;
Put u= 0, s=30 m and .
t=6.45 s
Calculate the time taken to complete his race.
T= t+t'
Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.
Put s= 30 m, and s'= 100 m.
T= 17.20 s
Therefore, the time taken for the race is 17.20 s.