The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif
1 answer:
Since the electric field between the plates is constant
so we can say the relation between potential difference and electric field is given by
here from above relation we can say that is electric field is kept constant then potential difference is directly proportional to the distance between the plates
so here we will say that if the distance between the plates is decreased then the potential difference between the plates will also decreases
So correct answer will be
<u>a) decrease </u>
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Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
<em>Where</em>:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
<em>Where:</em>
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
Now, we can find the work:
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
Answer:
The electric field is
Explanation:
Given that,
Radius = 2.00 cm
Number of turns per unit length
Current
We need to calculate the induced emf
Where, n = number of turns per unit length
A = area of cross section
=rate of current
Formula of electric field is defined as,
Where, r = radius
Put the value of emf in equation (I)
....(II)
We need to calculate the rate of current
....(III)
On differentiating equation (III)
Now, put the value of rate of current in equation (II)
Hence, The electric field is