Explanation:
The magnitude of the electric field between the plates is given by
E = -ΔV/d
minus sign indicates Potential decreases in the direction of electric field
where
ΔV is the potential difference between the plates
D is the distance between the plates.
The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface. That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.
The acceleration due to gravity near the surface of the planet is 27.38 m/s².
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Acceleration due to gravity near the surface of the planet</h3>
g = GM/R²
where;
- G is universal gravitation constant
- M is mass of the planet
- R is radius of the planet
- g is acceleration due to gravity = ?
g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²
g = 27.38 m/s²
Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².
Learn more about acceleration due to gravity here: brainly.com/question/88039
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1). the product of the two masses being gravitationally attracted to each other
2). the distance between their centers of mass
And that's IT. The gravitational force between them depends on
only those two things, nothing else.
Explanation:
hope this helps you dear friend.
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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