A particle with a mass of 2.0×10−5 kg and a charge of +2.0 microC is released in a (parallel plate) uniform horizontal electric field of 13 N/C. If the plates are 5.6 cm on each side, how much charge is on each? (in pC)
From what I have so far, the formula E=(4piKQ)/A; E=12 N/C, K=9x10^9 (Nm^2)/C^2, A=(0.056m)^2.
Whenever I plug this into my calculator, the answer I get for Q=3.6x10^-13 C, and the final answer to be 3.6x10^-22 pC.
Am I doing the arithmetic wrong? Any help would be greatly appreciated.
To find the charge on each plate, you can use the formula for electric field strength between parallel plates:
E = σ / (2ε₀)
Where E is the electric field strength, σ is the surface charge density on each plate, and ε₀ is the permittivity of free space.
We can rearrange this equation to solve for σ:
σ = 2ε₀E
Now, we need to find the value of ε₀. The permittivity of free space is a constant value equal to approximately 8.85 x 10^-12 C²/Nm².
Next, let's plug in the given values:
E = 13 N/C
ε₀ = 8.85 x 10^-12 C²/Nm²
σ = 2 * (8.85 x 10^-12 C²/Nm²) * (13 N/C)
= 2 * 8.85 x 10^-12 C
= 17.7 x 10^-12 C
Now, we can find the total charge on each plate by multiplying the surface charge density by the area of each plate. The area of each plate is given as (5.6 cm)^2. However, it would be better to convert this to meters first, so let's convert it:
(5.6 cm)^2 = (5.6 x 10^-2 m)^2
= 31.36 x 10^-4 m²
Now, let's find the total charge Q for each plate:
Q = σ * A
= (17.7 x 10^-12 C) * (31.36 x 10^-4 m²)
= 555.072 x 10^-16 C
= 555.072 x 10^-16 x 10^12 pC (converting to pC for the final answer)
= 555.072 x 10^-4 pC
≈ 5.55 x 10^-1 pC
So, the charge on each plate is approximately 5.55 x 10^-1 pC. It seems there was a calculation error in your previous work.