1.) a charge of -2x10^-9 C in an electric field between two metal plates 4cm apart is acted upon by a force of 10^-4 N. a.) what is the strength of the field? b.) what is the potential difference between the plates?
my question: how will going to solve this kind of problem? what do it mean "what is the strength of field?" and "what is the potential difference between the plates?"
pls help me on here.. tnx
Potential between plates is the voltage.
strength of field? What is the electric field strength. If you do not know what I am saying, you need a tutor immediately.
F/q=E solve for E.
V/d=E solve for V.
Well, well, well, looks like we have a shocking question about electric fields and potential differences! Don't worry, I'm here to help you zap through it with some clownish humor.
First things first, let's break it down. The strength of the electric field refers to how forceful the field is pushing on the charged particle. It's like asking how strong a breeze is on a windy day. You want to know how hard it's pushing, and in which direction.
Secondly, the potential difference between the plates is like asking about the difference in height in a roller coaster ride. It tells us how much energy is gained or lost by a charged particle when it moves between the plates. It's measured in volts, just like how you measure the thrill of a roller coaster ride in screams-per-minute.
Now, onto solving the problem! We need to use some equations here:
a.) To find the strength of the electric field, we can use the formula:
Electric Field Strength = Force / Charge
So, plug in the given numbers:
Electric Field Strength = 10^-4 N / -2x10^-9 C
Calculate it, and you'll get the answer in N/C, which represents the strength of the electric field.
b.) To find the potential difference between the plates, we can use another formula:
Potential Difference = Electric Field Strength * Distance
Here, the distance between the plates is given as 4 cm, so convert it to meters:
Potential Difference = Electric Field Strength * (4 cm * 0.01 m/cm)
And calculate it to get the answer in volts.
So, there you go! You've been zapped with the knowledge of how to solve the problem. Just remember, if you ever get stuck, just call on Clown Bot for some electrically charged humor. Good luck, my electrifying friend!
To solve this problem, we can use the formulas related to electric fields and potential difference.
a) "Strength of the field" refers to the electric field strength. The electric field strength is defined as the force experienced by a unit positive charge placed in the field. It is denoted by the symbol "E" and is measured in Newtons per Coulomb (N/C).
b) "Potential difference between the plates" refers to the voltage or electric potential difference between the two metal plates. It is a measure of the electric potential energy per unit charge required to move a charge from one plate to the other. It is denoted by the symbol "V" and is measured in Volts (V).
Now, let's solve the problem step-by-step:
Step 1: Determine the electric field strength (E).
- The formula for electric field strength is E = F/q, where F is the force acting on the charge and q is the magnitude of the charge.
- In this case, the charge is -2x10^-9C and the force is 10^-4 N.
- Plugging these values into the formula, we get E = (10^-4 N)/(-2x10^-9 C).
Step 2: Calculate the value of electric field strength (E).
- E = (10^-4 N)/(-2x10^-9 C) = -5x10^4 N/C (note the negative sign represents the direction of the field).
Step 3: Determine the potential difference (V).
- The formula for potential difference is V = E*d, where E is the electric field strength and d is the distance between the plates.
- In this case, the distance between the plates is given as 4 cm, which is equivalent to 0.04 m.
- Plugging in the values, we have V = (-5x10^4 N/C)*(0.04 m).
Step 4: Calculate the value of potential difference (V).
- V = (-5x10^4 N/C)*(0.04 m) = -2x10^3 V (note the negative sign represents the direction of the potential difference).
Therefore, the answers to the questions are:
a) The strength of the field is -5x10^4 N/C.
b) The potential difference between the plates is -2x10^3 V.
It is important to note that the negative signs indicate the direction of the field and potential difference, but do not affect the magnitude of the values.
To solve this problem, you will need to use the formulas related to electric fields and potential difference. Let's break down the problem into two parts:
a) What is the strength of the electric field?
The electric field strength (E) is the force (F) experienced by a charged object (Q) per unit charge. In this case, the force acting on the charge is given as 10^-4 N, and the charge is -2x10^-9 C. The formula is:
E = F / Q
Substituting the given values, you can calculate the electric field strength.
b) What is the potential difference between the plates?
The potential difference (ΔV) between two points is the work (W) required to move a unit positive charge (q) between those points. The formula to calculate potential difference is:
ΔV = W / q
In this case, since the charge is negative and the plates are metal, the potential difference is found to be the same as the magnitude of the electric field. So, the potential difference is equal to the electric field strength calculated in part a.
Now, let's calculate the answers:
a) To find the strength of the electric field:
E = F / Q
E = (10^-4 N) / (-2x10^-9 C)
E = -5x10^4 N/C (Note: the negative sign indicates that the electric field is in the opposite direction to the force acting on the charge)
b) To find the potential difference between the plates:
ΔV = E (same as the magnitude of the electric field)
ΔV = 5x10^4 V (Note: since potential difference is a scalar value, no negative sign is present)
Therefore, the strength of the electric field is -5x10^4 N/C, and the potential difference between the plates is 5x10^4 V.