A -0.06 charge that moves downward is in a uniform electric field with a strengthened of 200 N/C. What is the magnitude and dire
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As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards
So here we know that
now from the above equation
so both of the charges will apply 0.288 N force on q3 charge along the line joining them
now the net force due to vector sum is given by
here we know that angle is
now we have
so net force on q3 is 0.46 N vertically upwards along +Y axis
Answer:
1408.685 KN/C
Explanation:
Given:
R = 0.45 m
σ = 175 μC/m²
P is located a distance a = 0.75 m
k = 8.99*10^9
- The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:
part a)
Electric Field strength at point P: a = 0.75 m
part b)
Since, R >> a, we can approximate a / R = 0 ,
Hence, E simplified relation becomes:
E = σ / 2*e_o
part c)
Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:
Electric Field strength due to point charge is:
E = k*δ*pi*R^2 / a^2
Since, R << a, Surface area = δ*pi
Hence,
E = (k*δ*pi/a^2)