The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
<h3>Electric field on the master charge</h3>
E = kq/r²
where;
- q is magnitude of master charge
- r is distance of separation
- k is Coulomb's constant
E = (9 x 10⁹ x 0.63)/(0.75²)
E = 1.008 x 10¹⁰ N/C
<h3>Force on the test charge</h3>
F = Eq
where;
- E is electric field
- q is the test charge
F = (1.008 x 10¹⁰) x (0.5)
F = 5.04 x 10⁹ N
Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
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Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.
Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.
Case 1: Terminal velocity of a piece of tissue paper.
The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.
Downward gravitational force, F = mg
Upward air resistance or friction or drag force will be 
So, paper will attain terminal velocity when mg =
Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.
Downward force on rock, F = Mg
Drag force =
Rock will attain terminal velocity when Mg =
Mg > mg
so,
>
And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.