Answer:
1.7 J
Explanation:
The energy carried by a single photon is given by
E=hf
where h is the Planck's constant and f is the frequency of the photon.
The photon of our exercise has a frequency of f=1.7 \cdot 10^{17} Hz, therefore its energy is
E=hf=(6.63 \cdot 10^{-34}Js)(1.7 \cdot 10^{17} Hz)=1.1 \cdot 10^{-16} J
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Given:
The magnitude of each charge is q1 = q2 = 1 C
The distance between them is r = 1 m
To find the force when distance is doubled.
Explanation:
The new distance is
The force can be calculated by the formula
Here, k is the constant whose value is
On substituting the values, the force will be
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 )
