Answer:
Explanation:
The magnitude of the electrostatic force between two charges is given by Coulomb's law:
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
In this problem, at the beginning we have:
is the first charge
is the second charge
r is their initial separation
So the initial force is
Later, we have:
qA is doubled and r is doubled
This means that:
is the new charge
is the new separation
So the new force is:
Answer:
Explanation:
Given that there are two force of 1 pound each at right angles to each other.
The from the vector law of addition:
where:
resultant force
be the two of the forces to be added.
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
