The electric potential at the origin of the xy coordinate system is negative infinity
<h3>What is the electric field due to the 4.0 μC charge?</h3>
The electric field due to the 4.0 μC charge is E = kq/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q = 4.0 μC = 4.0 × 10 C and
- r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m
<h3>What is the electric field due to the -4.0 μC charge?</h3>
The electric field due to the -4.0 μC charge is E = kq'/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q' = -4.0 μC = -4.0 × 10 C and
- r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m
Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is
E" = E + E'
= -2E
= -2kq/r²
<h3>What is the electric potential at the origin?</h3>
So, the electric potential at the origin is V = -∫₂⁰E".dr
= -∫₂⁰-2kq/r².dr
Since E and dr = dx are parallel and r = x, we have
= -∫₂⁰-2kqdxcos0/x²
= 2kq∫₂⁰dx/x²
= 2kq[-1/x]₂⁰
= -2kq[1/x]₂⁰
= -2kq[1/0 - 1/2]
= -2kq[∞ - 1/2]
= -2kq[∞]
= -∞
So, the electric potential at the origin of the xy coordinate system is negative infinity
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Explanation:
it is almost zero .this is because the distance and the electrostatic force are inversely proportional
To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.
By Hook's law we know that force is defined as,
Where,
k = spring constant
x = Displacement change
PART A) For the case of the spring constant we can use the above equation and clear k so that
Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m
PART B) In the case of speed we can obtain it through the period, which is given by
Re-arrange to find \omega,
Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to
Therefore the mass of the trailer is 4093.55Kg
PART C) The frequency by definition is inversely to the period therefore
Therefore the frequency of the oscillation is 0.4672 Hz
PART D) The time it takes to make the route 10 times would be 10 times the period, that is
Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s