Answer:
1.02*10^-5 C/m²
Explanation:
Given that
Radius of the smaller sphere, r = 0.05 m
Radius of the larger sphere, R = 0.12 m
Electric field, E = 2*10^5 V/m
Formula for the electric field is
E = Q/(4πεR²)
this then means that the surface charge density of the larger sphere is
Q/4πR² = Eε = 1.77*10^-6 C/m²
and
Q = 4πεER² = 3.203*10^-7 C
is the charge on the large sphere, which is the same as the charge on the small sphere since they are connected by the wire
so the surface charge density of the smaller sphere is
Q/4πr² = 4πεER²/4πr²
Q = εER²/r²
Q = 1.02*10^-5 C/m²
Answer:
98 N
Explanation:
Given that The weight of an object is the product of its mass, m, and the acceleration of gravity, g (where g=9.8 m/s2). Of an object’s mass is m=10. kg,
The parameters to be considered are
Mass m = 10 kg
Acceleration due to gravity = 9.8m/s^2
Weight W = mg
substitute mass m and acceleration due to gravity g into the formula above.
Weight = 10 × 9.8
Weight = 98 N
Therefore, the weight of the body or object is 98 N
Answer:
-1.4N
Explanation:
When the distance between A and B is 13.8mm (0.0138m), the force exerted by A on B is 2.65N:
F(A,B) = k*qA*qB/r²
=> 2.65 = k*qA*qB / 0.0138²
=> k*qA*qB = 2.65 * 0.0138²
k*qA*qB = 0.0005046
When the distance between them increases to 19mm (0.019m), the force exerted by A on B becomes:
F(A,B) = k*qA*qB / 0.019²
Given that k*qA*qB = 0.0005046,
F(A,B) = 0.0005046/0.019²
F(A,B) = 1.4N
The force exerted by B on A will be in the opposite direction of the force exerted by A on B, which means that the force exerted by B on A will be:
F(B, A) = -1.4N