Answer:
The potential is [tex]V_A = 9600 \ V[/tex]
Explanation:
From the question we are told that
The magnitude of the charge is [tex]q_1 = 4 \mu C = 4*10^{-6} \ C[/tex]
The position of the charge is [tex]x = + 3.0 \ m[/tex]
The magnitude of the second charge is [tex]q_2 = -2.0 \mu C = -2.0 *10^{-6} \ C[/tex]
The position is [tex]y_1 = - 1.0 \ m[/tex]
The position of point A is [tex]y_2 = + 4.0 \ m[/tex]
Generally the electric potential at A due to the first charge is mathematically represented as
[tex]V_a = \frac{k * q_1 }{r_1 }[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]
[tex]r_1[/tex] is the distance between first charge and a which is mathematically represented as
[tex]r_1 = \sqrt{x^2 + y_2 ^2 }[/tex]
=> [tex]r_1 = \sqrt{3^2 + 4 ^2 }[/tex]
=> [tex]r_1 = 5 \ m[/tex]
So
[tex]V_a = \frac{9*10^9 * 4*10^{-6} }{5 }[/tex]
[tex]V_a = 7200 \ V[/tex]
Generally the electric potential at A due to the second charge is mathematically represented as
[tex]V_b = \frac{k * q_2 }{r_2 }[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]
[tex]r_2[/tex] is the distance between second charge and a which is mathematically represented as
[tex]r_2 = y_2 - y[/tex]
=> [tex]r _2 = 4.0 - (-1.0)[/tex]
=> [tex]r = 5 \ m[/tex]
So
[tex]V_a = \frac{9*10^9 * -2*10^{-6} }{5 }[/tex]
[tex]V_a = -3600 \ V[/tex]
So the net potential difference at point A due to the charges is mathematically represented as
[tex]V_n = V_a + V_b[/tex]
=> [tex]V_n = 7200 - 3600[/tex]
=> [tex]V_n = 3600 V[/tex]
Generally the net potential difference at the origin due to both charges is mathematically represented as
[tex]V_N = V_c + V_d[/tex]
Here
[tex]V_c = \frac{k * q_1 }{x}[/tex]
=> [tex]V_c = \frac{9*10^9 * 4*10^{-6} }{3}[/tex]
=> [tex]V_c = 12000 V[/tex]
and
[tex]V_d= \frac{k * q_2 }{y}[/tex]
=> [tex]V_c = \frac{9*10^9 * -2*10^{-6} }{1}[/tex]
=> [tex]V_c =- 18000 V[/tex]
Generally the net potential difference at the origin is
[tex]V_N = 12000 - 18000[/tex]
=> [tex]V_N = -6000[/tex]
Generally the potential difference at A relative to zero at the origin is mathematically evaluated as
[tex]V_A = V_n - V_N[/tex]
=> [tex]V_A = 3600 - (-6000)[/tex]
=> [tex]V_A = 9600 \ V[/tex]
Conduction In Everyday Life Examples
at least 5 plz help
1. Heat transfer from a hot burner on the stove into a pot or pan
2. A metal spoon that gets hot from the boiling water
3. Chocolate candy in your hand that melts because of the heat
4. The wires in your house that conduct electricity
5. Roasting marshmallows, the heat will soften the marshmallow
Answer:
the engine heat from a car you just turned on.
a heater
a pad warmer
a hot pot
using a campfire to roast hot dogs