1. what happens to the field strength when you halve the distance between two charges?
2. A positive charge of 3.5 uC is located in an electric field of intensity 3.00 N/C directed toward the south. Find the magnitude and the direction of the force on the test charge.
3. Three resistors of 10.0 ohms, 20.0 ohms, and 25.0 ohms are connected in patallel across a 100-V battery. What is the equivalent resistance of the circuit and the current flowing through the 10.0-ohm resistor?
(1) electric field strength E =k•q/r²
If r1 =r/2,
then E1 =4•E.
(2)
F =q•E
(3)
1/R =1/R1 +1/R2 + 1/R3.
R =R1•R2•R3/(R2•R3 + R1•R3 + R1•R2)
U =I1•R1,
I1 =U/R1
thanks so much!
1. When you halve the distance between two charges, the field strength at that point increases by a factor of 4.
To understand why this happens, we need to look at the equation for the electric field strength between two point charges, which is given by:
E = k * (Q / r^2)
Where E is the electric field strength, k is the electrostatic constant, Q is the charge, and r is the distance between the charges.
When you halve the distance (r), you are effectively squaring it (r^2). Since r^2 appears in the denominator of the equation, when you decrease its value by half, the fraction (1/r^2) becomes larger, resulting in a larger electric field strength (E).
2. To find the magnitude and direction of the force on a test charge in an electric field, you can use the equation:
F = q * E
Where F is the force, q is the charge of the test charge, and E is the electric field intensity.
In this case, the charge of the test charge (q) is given as 3.5 uC (microcoulombs) and the electric field intensity (E) is given as 3.00 N/C directed toward the south.
To find the magnitude of the force, simply multiply the charge and the electric field intensity:
F = (3.5 * 10^-6 C) * (3.00 N/C) = 10.5 * 10^-6 N = 10.5 uN
The direction of the force is the same as the direction of the electric field, which is toward the south.
So, the magnitude of the force is 10.5 microNewtons and it is directed toward the south.
3. To find the equivalent resistance of a parallel circuit with three resistors and the current flowing through a specific resistor, you can use the following formulas:
For the equivalent resistance (Req) of resistors connected in parallel, the equation is:
1/Req = 1/R1 + 1/R2 + 1/R3
Where R1, R2, and R3 are the resistances of the individual resistors.
In this case, R1 = 10.0 ohms, R2 = 20.0 ohms, and R3 = 25.0 ohms. Plugging these values into the equation, we get:
1/Req = 1/10 + 1/20 + 1/25 = 0.1 + 0.05 + 0.04 = 0.19
To find Req, we take the reciprocal of both sides of the equation:
Req = 1/0.19 = 5.26 ohms
So, the equivalent resistance of the circuit is 5.26 ohms.
To find the current flowing through the 10.0-ohm resistor (I1), we can use Ohm's Law:
I = V / R
Where I is the current, V is the voltage, and R is the resistance.
In this case, the voltage (V) is given as 100 V and the resistance (R) is 10.0 ohms.
Plugging these values into the equation, we get:
I1 = 100 V / 10.0 ohms = 10 A
So, the current flowing through the 10.0-ohm resistor is 10 Amperes.