What is the electric potential difference of 5cm from a point charge of -2.0µc
1 answer:
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Answer:
1. 24375 N/C
2. 2925 V
Explanation:
d = 12 cm = 0.12 m
F = 3.9 x 10^-15 N
q = 1.6 x 10^-19 C
1. The relation between the electric field and the charge is given by
F = q E
So,
E = 24375 N/C
2. The potential difference and the electric field is related by the given relation.
V = E x d
where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.
By substituting the values, we get
V = 24375 x 0.12 = 2925 Volt
Answer:
1 x 10¹⁷
Explanation:
Given data:
Radius of the earth = 6000km
Radius of an atom = 60pm
Now, how many orders is the radius of the earth larger than an atom
Solution:
To solve this problem, let us express both quantity as the same unit;
1000m = 1km
6000km = 6000 x 10³m = 6 x 10⁶m
60pm;
1 x 10⁻¹²m = 1pm
60pm = 60 x 1 x 10⁻¹²m = 6 x 10⁻¹¹m
Now;
The order: = 1 x 10¹⁷
(a) 0.96 m/s
The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:
The frequency of the waves is the reciprocal of the period:
The wavelength instead is just the distance between two consecutive crests, so
And the wave speed is given by:
(b) 0.265 m
The total distance between the highest point of the wave and its lowest point is
d = 0.53 m
The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is
(c) Amplitude: 0.15 m, wave speed: same as before
In this case, the amplitude of the wave would be lower. In fact,
d = 0.30 m
So the amplitude would be
Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.