Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining them is the electric field zero?
(a) 15 cm from charge 4q
(b) 20 cm from charge 4q
(c) 7.5 cm from charge q
(d) 5 cm from charge q
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To find the point on the line joining the charges where the electric field is zero, we can make use of Coulomb's Law. Coulomb's Law states that the electric field created by a point charge is given by:
E = k * q / r^2
Where:
E is the electric field,
k is the electrostatic constant (approximately equal to 9 x 10^9 Nm^2/C^2),
q is the charge, and
r is the distance from the charge to the point where we want to calculate the electric field.
In this case, we have two charges, +4q and +q, placed 30 cm apart. Let's assume that we want to find the point on the line where the electric field is zero. We can call this point "P".
Let's calculate the electric field at point P due to the +4q charge. We'll use the equation:
E1 = k * (4q) / (30 - x)^2
Where x represents the distance from the +4q charge to point P.
Similarly, let's calculate the electric field at point P due to the +q charge:
E2 = k * q / x^2
Since we want the total electric field to be zero at point P, we can set up the equation:
E1 + E2 = 0
Substituting the equations for E1 and E2, we get:
k * (4q) / (30 - x)^2 + k * q / x^2 = 0
Simplifying this equation, we get:
4 / (30 - x)^2 + 1 / x^2 = 0
To solve this equation, we can find a common denominator:
(4 * x^2 + (30 - x)^2) / (x^2 * (30 - x)^2) = 0
Expanding the numerator, we get:
(4x^2 + 900 - 60x + x^2) / (x^2 * (30 - x)^2) = 0
Simplifying further, we get:
(5x^2 - 60x + 900) / (x^2 * (30 - x)^2) = 0
Since the numerator is a quadratic expression, for the fraction to be zero, the numerator must be equal to zero:
5x^2 - 60x + 900 = 0
Simplifying this quadratic equation further, we can divide the equation by 5 to simplify the coefficients:
x^2 - 12x + 180 = 0
Now, we can solve this quadratic equation to find the values of x. Factoring or using the quadratic formula, we get:
(x - 6)(x - 30) = 0
The solutions for x are x = 6 and x = 30.
Now, let's analyze the possible answer choices:
(a) 15 cm from charge 4q: This is halfway between the two charges. Electric field is not zero at this position.
(b) 20 cm from charge 4q: This is closer to the +4q charge. Electric field is not zero at this position.
(c) 7.5 cm from charge q: This is 7.5 cm from the +q charge. Electric field is not zero at this position.
(d) 5 cm from charge q: This is 5 cm from the +q charge. We found one of the solutions to be x = 6, which is close to 5 cm. Therefore, the electric field should be zero at this position.
Hence, the correct answer is (d) 5 cm from charge q.
The distance to the +q charge must be half the distance to the +4q charge, so that the electric fields cancel. Remember the inverse square dependence upon distance!
That puts you 10 cm from q and 20 cm from 4q.
That happens to be one of your choices.