A +50 micro coulomb and 30 micro coulomb charge are placed 50cm apart.
a)what is the direction and magnitude of thee electric field at a point P, in between them, that is 10cm from the 30 micro coulomb charge?
b)if an electron is placed at rest P, what will its acceleration be initially?
c)at what points along the line joining them is the electric field zero?
d)at what points along the line joining them is the potential zero?
q1…………….<----- ------ x -->….q2
E2 P E1
E = k•q/r^2
E1 = k•q1/r1^2 = 9•10^9•50•10^-6/(0.4)^2 = 2,8•10^6 V/m,
E2 = k•q2/r2^2 = 9•10^9•30•10^-6/(0.1)^2 = 2.7•10^7 V/m,
E =E2 – E1 = 2,42•10^7 V/m.
(b)
ma =eE,
a =eE/m = 1.6•10^-19•2.42•10^7/9. 1•10^-31 =4.26•10^18 m/s².
(c)
x distance from the 50 micro coulomb)
k•q1/x²= k•q2/(0.5 –x)²
5/x² =4/(0.5–x)²
Solve for x.
(d)
k•q1/x = k•q2/(0.5 –x)
5/x =4/(0.5–x)
Solve for x
What is the magnitude of the force exerted on q1 with 10 micro-Coulombs charge by q2 with a charge of -20 micro-Coulombs if their separating distance is 50 cm?
To answer these questions, we will use Coulomb's Law and the concept of electric fields. Coulomb's Law states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
a) To find the electric field at point P, we need to calculate the individual electric fields due to each charge and then add them vectorially. The electric field due to each charge at point P can be calculated using Coulomb's Law:
Electric field due to charge Q1 = (k * Q1) / r1^2
Electric field due to charge Q2 = (k * Q2) / r2^2
Where Q1 and Q2 are the charges, r1 and r2 are the distances from the respective charges to point P, and k is the electrostatic constant (9 × 10^9 Nm^2/C^2).
Given:
Charge Q1 = +50 μC = 50 × 10^-6 C
Charge Q2 = +30 μC = 30 × 10^-6 C
Distance r1 = 10 cm = 0.1 m
Distance r2 = 50 cm = 0.5 m
Substituting the values into the formula, we get:
Electric field due to Q1 = (9 × 10^9 Nm^2/C^2) * (50 × 10^-6 C) / (0.1 m)^2
Electric field due to Q2 = (9 × 10^9 Nm^2/C^2) * (30 × 10^-6 C) / (0.5 m)^2
Calculate these values to find the magnitudes of the electric fields. Since the charges are positive, the electric fields will act radially outward from each charge. To find the resultant electric field at point P, add these two vectors using vector addition.
b) The initial acceleration of an electron at point P can be calculated using the formula:
F = m * a
Where F is the force, m is the mass, and a is the acceleration.
The force experienced by the electron can be calculated using Coulomb's Law:
F = k * (|Q1| * |Qe|) / r^2
Given:
Charge Qe = -1.6 × 10^-19 C (charge of an electron)
Distance r = 10 cm = 0.1 m
Substituting the values into the formula, we get:
F = (9 × 10^9 Nm^2/C^2) * (50 × 10^-6 C) * (1.6 × 10^-19 C) / (0.1 m)^2
Once we have the force, we can calculate the acceleration using F = m * a. The mass of an electron is approximately 9.11 × 10^-31 kg.
c) The electric field along the line joining the two charges will be zero at points where the magnitude and direction of the electric fields due to the two charges cancel each other out. Mathematically, this occurs when E1 = -E2, where E1 is the electric field due to charge Q1 and E2 is the electric field due to charge Q2. Calculate these electric fields for different positions along the line and check where E1 and E2 are equal in magnitude and opposite in direction.
d) The potential at a point is defined as the work done per unit positive charge in bringing a test charge from infinity to that point. The potential due to a point charge can be calculated using the formula:
V = (k * Q) / r
Where V is the potential, Q is the charge, r is the distance from the charge to the point, and k is the electrostatic constant (9 × 10^9 Nm^2/C^2). Along the line joining the two charges, the potential will be zero at points where the potentials due to the two charges cancel out. Calculate the potential due to each charge for different positions along the line and check where the potentials are equal in magnitude but opposite in sign (one positive, one negative).