Two electrostatic point charges of +50.0 μC
and +50.0 μC exert a repulsive force on each
other of 197 N.
What is the distance between the two
charges? The Coulomb constant is 8.98755 ×
109 N · m2/C2.
Answer in units of m.
From electric field force b/w two charges=F=Q1Q2/4pieE¤r (square)
force(f)=197,Q1=50,Q2=50,and 4pie E¤r(square)=8.98755*109Nm/C
:->
197=50*50 /8.98755*109*r
197*8.98755*109*r2=50.0*50.0
r2=192730.912/2500
after division then take d square root of d ans. Given to get ur( r) as d distance.
To calculate the distance between two charges, we can use Coulomb's Law:
F = k * (q1 * q2) / r^2,
where F is the force between the charges, k is the Coulomb constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
Given:
F = 197 N,
k = 8.98755 × 10^9 N · m^2/C^2,
q1 = +50.0 μC = 50.0 × 10^-6 C,
q2 = +50.0 μC = 50.0 × 10^-6 C.
We need to rearrange the equation to solve for r:
r^2 = (k * (q1 * q2)) / F.
Substituting the given values:
r^2 = (8.98755 × 10^9 N · m^2/C^2 * (50.0 × 10^-6 C * 50.0 × 10^-6 C)) / 197 N.
Now we can calculate the value of r:
r^2 = (8.98755 × 10^9 N · m^2/C^2 * (2.5 × 10^-9 C^2)) / 197 N.
simplifying the expression gives:
r^2 = 3.632585177 N · m^2/C^2.
Taking the square root of both sides:
r = sqrt(3.632585177 N · m^2/C^2).
Calculating the square root:
r ≈ 1.903 m.
Therefore, the approximate distance between the two charges is 1.903 meters.