Answer:
1.6 x 10⁻¹⁹ C
Explanation:
Let us arrange the charges in the ascending order and round them off as follows :-
1.53 x 10⁻¹⁹ C → 1.6x 10⁻¹⁹ C
3.26 x 10⁻¹⁹C → 3.2 x 10⁻¹⁹ C
4.66 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C
5.09 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C
6.39 x 10⁻¹⁹C → 6.4 x 10⁻¹⁹ C
The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.
Here we observe that
2 nd charge is almost twice the first charge
3 rd and 4 th charges are almost 3 times the first charge
5 th charge is almost 4 times the first charge.
This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first charge , 2nd to 5 th charges can be written as 2e, 3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of 1.6 x 10⁻¹⁹ C exists.
Answer:
120 m/s
Explanation:
Given:
v₀ = 0 m/s
a = 12 m/s²
t = 10 s
Find: v
v = at + v₀
v = (12 m/s²) (10 s) + 0 m/s
v = 120 m/s
The distance between the two charges is 
Explanation:
The electrostatic force between two charged objects is given by Coulomb's law:

where:
is the Coulomb's constant
are the charges of the two objects
r is the separation between the two charges
In this problem, we are given the following:



Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

Learn more about electrostatic force:
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Answer:
Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²
Explanation:
Given;
power of radio transmitter, P = 63.2 kW = 63200 W
distance of transmission, r = 30.5 km
Intensity of the transmitted radio wave is calculated as follows;

where;
I is the intensity of the transmitted radio wave
Substitute the given values and calculate the intensity of the transmitted radio wave;

Therefore, Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²