The s<span>tatic charges that are applied to neutral objects by friction, induction or conduction main utilise the creation of electrons or electric charges. Electrons flow forming a current into a specific medium known as a conductor which is mainly due to a significant potential difference between two points.</span>
Answer:
1.-E=1000N/C to the LEFT
2.-The electric field inside a conductor in electrostatic state is always zero (conductor proprieties).
3.-The voltmeter read 0V as differential voltage between two points from the conductor
Explanation:
1.The electric field inside the conductor must be zero (conductor proprieties). Then the charges create a electric field equal an opposite to the external electric field. In other words E=1000N/C to the LEFT
2. The electric field inside a conductor in electrostatic state is always zero. As shown in the figure the electric field induced by the charges in the sphere surface cancelled the EXTERN electric field.
3.If the Electric field inside the conductor is zero, that means that the Voltage in the hole conductor is constant (conductor proprieties). In other words the the voltmeter read 0v as differential voltage between two points from the conductor.
Answer:
0.36 A.
Explanation:
We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:
Resistor 1 (R₁) = 35 Ω
Resistor 2 (R₂) = 20 Ω
Equivalent Resistance (Rₑq) =?
Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:
Rₑq = (R₁ × R₂) / (R₁ + R₂)
Rₑq = (35 × 20) / (35 + 20)
Rₑq = 700 / 55
Rₑq = 12.73 Ω
Next, we shall determine the total resistance in the circuit. This can be obtained as follow:
Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω
Resistor 3 (R₃) = 15 Ω
Total resistance (R) in the circuit =?
R = Rₑq + R₃ (they are in series connection)
R = 12.73 + 15
R = 27.73 Ω
Finally, we shall determine the current. This can be obtained as follow:
Total resistance (R) = 27.73 Ω
Voltage (V) = 10 V
Current (I) =?
V = IR
10 = I × 27.73
Divide both side by 27.73
I = 10 / 27.73
I = 0.36 A
Therefore, the current is 0.36 A.
Answer:
8 units
Explanation:
F=(k*q1*q2)/(r^2)
K is a constant, q1 is charge of 1, q2 is charge of 2, r is distance between the two.
Answer:

Explanation:
Given data
Electric potential at point a is Ua=5.4×10⁻⁸J
q₂ moves to point b where a negative work done on it
Required
Electric potential energy Ub
Solution
When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

Now substitute the given values
So
