Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
distance "a" from the ring's center.
1 answer:
Answer:
E=
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by
Substitute x=a and R=a
Then, we get
Where K=
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=
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Answer:
D/H =15
Explanation:
- We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
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- If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:
- We can use the same equation, to find the value of D, as follows:
- In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
- When we replace these values in (1), we find that v₁ = -v₀.
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- From (2) we know that H can be expressed as follows:
Answer:
72
Explanation:
Formula
Explanation:
1. serie
2.Red
3.16 ohms
4.
I=V/R
6=V/16
V=16×6
V=96 volts
m= 60g = 60/1000 Kg = 0.06 Kg
v = 2cm3 = 2 * (0.01^3) m3 = 2 *10^-6 m3
Density= m/v = 6 * 10^-2 / 2 *10^-6 = 3 *10^4 Kg/m3