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Dafna1 [17]
4 years ago
15

An electric field of magnitude e is measured at a distance r from a point charge q. if the charge is doubled to 2q, and the elec

tric field is now measured at a distance of 2r from the charge, what is the value of the new electric field?
Physics
1 answer:
pav-90 [236]4 years ago
5 0
Electric field is proportional to the charge and inversely proportional to the square of distance:
E\propto\frac{Q}{r^{2}}
With charge 2q and distance 2r, the electric field is proportional to:
E_{q}\propto\frac{2q}{(2r)^{2}}=\frac{2q}{4r^{2}}=\frac{1}{2}\frac{q}{r^{2}}
The new electric field is half of that of the original measured field.
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The question is based on the principle of motion in one dimension and hence formulas of motion in one dimension can be applied.

It is given that donkey attains an acceleration of 1.6 m/s^2

The time taken to accelerate  to given speed is 6.7 seconds

We use the formula v=u + at to find the fastest speed

v is the final or maximum speed

u is the initial speed which in this case is 0 as the donkey is at rest

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v = u + at

v= 0 + 1.6 x 6.7

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Hence the donkey obtains the speed of 10.72 m/s

For further reference:

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