The electric field between plates is 4000V/m.
An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.
The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.
The voltage between points A and B is
V=E.d
E =V/d (uniform E- field only)
where d is the distance from A to B, or the distance between the plates.
Given:
distance d = 3 cm
voltage V = 120 V
Electric field E = V/d
E = 120 V / 3cm
E = 40 V / 1 cm [ 1 cm = 1/100 m ]
E = 4000 V/m.
Learn more about Electric field here:
brainly.com/question/8971780
#SPJ4
The potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules
<h3>
What is the energy in a capacitor?</h3>
The energy stored in a capacitor is an electrostatic potential energy.
It is related to the charge(Q) and voltage (V) between the capacitor plates.
It is represented as 'U'.
<h3>
How to determine the potential difference</h3>
Formula:
Potential difference, V is the ratio of the charge to the capacitance of a capacitor.
It is calculated using:
V = Q ÷ C
Where Q = charge 5 × 10∧-5C and C = capacitance 10∧-9
Substitute the values into the equation
Potential difference, V = 5 × 10∧-5 ÷ 10∧-9 = 5 × 10∧4 volts
<h3>
How to determine the energy stored</h3>
Formula:
Energy, U = 1 ÷ 2 (QV)
Where Q= charge and V = potential difference across the capacitor
Energy, U = 1 ÷ 2 ( 5 × 10∧-5 × 5 × 10∧4)
= 0.5 × 25 × 10∧-1
= 0.5 × 2.5
= 1. 25 Joules
Therefore, the potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules
Learn more about capacitance here:
brainly.com/question/14883923
#SPJ1
<h2>
Answer:</h2>
(a) 10N
<h2>
Explanation:</h2>
The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force 
is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F -
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F - = m x a
F - = 50 x 0
F - = 0
F = -------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force is opposing the movement of the box.
∑F = 1.5F -
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F - = m x a
1.5F - = 50 x 0.1
1.5F - = 5 ---------------------(iii)
<em>Substitute </em><em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
<em />
Larger mass creates a stronger pull
To solve this we assume
that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 =284.15 x 2.50 / 303.15
<span>V2 = 2.34 L</span>
