Answer:
b)
Explanation:
By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.
This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.
(As the positive charge would move away from positive charges and would be attracted by negative ones).
So, the combination of answers that is true is b) (positive, negative, positive).
Answer:
241.7 s
Explanation:
We are given that
Charge of particle=
Kinetic energy of particle=
Initial time=
Final potential difference=
We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.
We know that
Using the formula
Initial voltage=
Using the formula
Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
<span>Radius, the distance from the centre = 0.390
Electric field is equal to half of the magnitude. E2 = E / 2
Given
E1 = E2
E1 = k x Q / r^2
E2 = (k x Q / r2^2) / 2
Equating the both we get 2 x r^2 = r2^2
r2 = square root of (2 x r1^2) = square root of (2) x r = 1.414 x 0.390
r2 = 1.414 x 0.390 = 0.551 m</span>
Answer:
a c d
Explanation:
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