Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
Answer:
solar eclipse
Explanation:
because at that time, the moon completely covers the path and casts its shadow on earth because it is present between sun and earth's path. so, solar eclipse occurs.
hope it helps :)
Answer:
The intensity will be 1/9 as much.
Explanation:
The intensity of the light or any source is inversely related to the square of the distance.
Now according to the question the distance is increased by three times than,
Therefore,
Therefore the intensity will become 1/9 times to the initial intensity.
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance 
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q =
= 
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
This can be solve using the formula P = I^2 * Rwhere P is the powerI is the CurrentR is the resistanceP = I^2 * R
1/4 Watt = I^2 * 100 ohm solve for II^2 = 1/400 I = 0.05 amps then using the formula to solve for the voltage:V = I * RV = 0.05 amps * 100 ohms V = 5 volts
