Answer: Option (b) is the correct answer.
Explanation:
It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.
Then test charge to the right immediately after being released.
Therefore, the net force will be as follows.
F = 
= 
= 
F = > 0
Thus, we can conclude that the test charge move to the right immediately after being released.
Answer:
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Answer:
applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²
Answer:
The right solution is:
(a) 2.87 eV
(b) 1.4375 eV
Explanation:
Given:
Wavelength,
= 433 nm
Potential difference,
= 1.43 V
Now,
(a)
The energy of photon will be:
E = 
= 
or,
= 
= 
(b)
As we know,
⇒ 
By substituting the values, we get
⇒ 
⇒ 
or,
⇒ 
⇒ 
