Respuesta :
Answer:
1.) 1000 N/c
2.) 200000 N
3.) 60v
4.) 120000J
5.) 4899 m/s
Explanation:
Given that a 100V battery is connected to two oppositely charged plates that are 10cm apart.
1.) The magnitude of electric field strength is the ratio of potential difference per distance between the plates. That is,
Electric field = V/d
Electric field = 100/0.1
Electric field = 1000 wb
2.) Electric field strength is also the ratio of force per unit ccharge. That is,
Electric field = F / q
1000 = F / 200
Cross multiply
F = 1000 × 200
F = 200000 N
3.) What is the electric potential energy of the charge when it's 8cm and 2cm from the negatively charged plate?
The distance = 8cm - 2cm = 6cm
Electric field = V / d
Substitute the electric field and the distance into the formula
1000 = V / 0.06
Make V the subject of formula
V = 1000 × 0.06
V = 60 V
4.) How much work was required to move the charge from 8cm to 2cm?
Work done = VQ
Since V = 60v
Then,
Work done = 60 × 200
Work done = 120000 J
5.) If the 10 gram point charge was at rest at point A, what is the final speed at point B?
Work done = 1/2mv^2
120000 = 1/2 × 0.01 × v^2
V^2 = 120000/ 0.005
V^2 = 24000000
V = 4899 m/s
The magnitude of electric field strength has been 1000 wb.
The electric force (F) applied by 200 C charge has been 2000,000 N.
The values for electric potential energy has been 60 V.
The work done in moving the charge across the plate has been 12 kJ.
The speed of charge at point B has been 48.98 m/s.
The magnitude of electric field strength has been defined as the force per unit charge.
- The expression for magnitude of electric field strength (E) has been:
[tex]E=\dfrac{V}{d}[/tex]
Where, the voltage of battery, [tex]V=100\;\rm V[/tex]
The distance between plates, [tex]d=10\;\rm cm\;;0.1\;m[/tex]
Substituting the values for magnitude of electric field strength (E):
[tex]E=\dfrac{100}{0.1} \\E=1000\;\rm wb[/tex]
The magnitude of electric field strength has been 1000 wb.
- The expression of electric force (F) applied by the battery has been given as:
[tex]F=E\;\times\;q[/tex]
Where, the magnitude of electric field strength, [tex]E=1000\;\rm wb[/tex]
The charge for the force has been, [tex]q=200\;\rm C[/tex]
Substituting the values for electric force (F):
[tex]F=1000\;\times\;200\\F=200000\;\rm N[/tex]
The electric force (F) applied by 200 C charge has been 2000,000 N.
- The electric potential energy (E) for charge has been given as:
[tex]E=\dfrac{V}{d}[/tex]
The magnitude of electric field strength, [tex]E=1000\;\rm wb[/tex]
The distance between plates, [tex]d=8\rm \;cm-6\;cm\\d=2\;cm\;;0.02\;m[/tex]
Substituting the values for electric potential energy, (V):
[tex]1000=\dfrac{V}{0.0.06} \\V=1000\;\times\;0.06\\V=60\;\rm V[/tex]
The values for electric potential energy has been 60 V.
- The work done (W) to move a charge has been given by:
[tex]W=Vq[/tex]
Where, the electric potential energy of the battery has been, [tex]V=60\;\rm V[/tex]
The charge for the force has been, [tex]q=200\;\rm C[/tex]
Substituting the values for work done, (W):
[tex]W=60\;\times\;200\\W=12,000\;\text J\\W=12\;\rm kJ[/tex]
The work done in moving the charge across the plate has been 12 kJ.
- The speed (v) of 10 g charge has been given as:
[tex]W=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2W}{m} }[/tex]
Where, the work done on unit charge, [tex]W=12,000\;\rm J[/tex]
The mass of the charge, [tex]m=10\;g[/tex]
Substituting the values for calculating the speed of charge, (v):
[tex]v=\sqrt{\dfrac{2\;\times\;12,000}{10} } \\v=\sqrt{24,000}\\v=48.98\;\rm m/s[/tex]
The speed of charge at point B has been 48.98 m/s.
For more information about electric field, refer to the link:
https://brainly.com/question/12757739
