The last one. Opposites attract while similar ones repulse each other so all forces are pushing them apart
The electric potential between the two charges is 91.68 V.
<h3>
Electric potential between the two charges</h3>
The electric potential between the two charges is calculated as follows;
V = Ed
where;
- V is electric potential
- E is electric field
- d is the distance of the charge
Substitute the given parameters and solve for electric potential,
V = 573 N/c x 0.16 m
V = 91.68 V
Thus, the electric potential between the two charges is 91.68 V.
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Answer:
Answer:
Q_1 = 7Q
1
=7
Q_2 = 10Q
2
=10
Q_3 = 13.5Q
3
=13.5
Step-by-step explanation:
Given
5, 7, 7, 8, 10, 11, 12, 15, 17.
Required
Determine Q1, Q2 and Q3
The number of data is 9
Calculating Q1:
Q1 is calculated as:
Q_1 = \frac{1}{4}(N + 1)Q
1
=
4
1
(N+1)
Substitute 9 for N
Q_1 = \frac{1}{4}(9 + 1)Q
1
=
4
1
(9+1)
Q_1 = \frac{1}{4}*10Q
1
=
4
1
∗10
Q_1 = 2.5th\ itemQ
1
=2.5th item
This means that the Q1 is the mean of the 2nd and 3rd data.
So:
Q_1 = \frac{1}{2}(7+7)Q
1
=
2
1
(7+7)
Q_1 = \frac{1}{2}*14Q
1
=
2
1
∗14
Q_1 = 7Q
1
=7
Calculating Q2:
Q2 is calculated as:
Q_2 = \frac{1}{2}(N + 1)Q
2
=
2
1
(N+1)
Substitute 9 for N
Q_2 = \frac{1}{2}(9 + 1)Q
2
=
2
1
(9+1)
Q_2 = \frac{1}{2}*10Q
2
=
2
1
∗10
Q_2 = 5th\ itemQ
2
=5th item
Q_2 = 10Q
2
=10
Calculating Q3:
Q3 is calculated as:
Q_3 = \frac{3}{4}(N + 1)Q
3
=
4
3
(N+1)
Substitute 9 for N
Q_3 = \frac{3}{4}(9 + 1)Q
3
=
4
3
(9+1)
Q_3 = \frac{3}{4}*10Q
3
=
4
3
∗10
Q_3 = 7.5th\ itemQ
3
=7.5th item
This means that the Q3 is the mean of the 7th and 8th data.
So:
Q_3 = \frac{1}{2}(12+15)Q
3
=
2
1
(12+15)
Q_3 = \frac{1}{2}*27Q
3
=
2
1
∗27
Q_3 = 13.5Q
3
=13.5
