What happens to the magnitude of the electric field of a charged particle with c

Question

What happens to the magnitude of the electric field of a charged particle with charge Q (at a distance R from the particle) if we increase its charge by a factor of 4 and increase the distance by a factor of 2?
a) the magnitude increases by a factor of two.
b) the magnitude decreases by a factor of two.
c) the magnitude stays the same.
d) the magnitude stays the same but the direction of the field changes.
e) none of the above.
9) Why it is important to have water in meals which are taken to a microwaves oven to be warmed up?
a) Water is a fundamental molecule to everything we know.
b) Water has electric dipoles which oscillate with the microwaves’ electric field. These
oscillations generate heat and, therefore, warm up the meals.
c) Water does not have electric dipoles and, therefore, are suitable to avoid electrical
discharges due to the microwaves.
d) Water has electric dipoles which oscillate with the microwaves’ magnetic field. These
oscillations generate heat and, therefore, warm up the meals.
e) none of the options.

Explanation / Answer

Given

charged particle of charge Q

at a distance R the electtric field is say E1

if we increase the charge by a factor of 4, that is Q2 = 4Q

and the distance is increased by a factor of 2 ==> R2 = 2*R

what is E2 =?

we know that the electric field is E = kq/r^2

here E1 = k*Q/R^2

and E2 = k*Q2/R2^2

= k*4Q/(2R)^2

= k *4Q/4R^2

= kQ/R^2

so here E1 = E2 no change in field

the answer is the magnitude stays the same -------(optionc)

----------

9) from the given options actually none of the above is correct

infact the radio waves or micro waves can heat teh water having electric dipoles in the water

not the electric fiedl and mangetic field

the answer is e) none of the above

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