Calculating Potential Difference in a Uniform Electric Field

AI Thread Summary
To calculate the potential difference in a uniform electric field, the correct approach involves using the formula ∆V = E(Sf - Si), where E is the electric field strength and Sf and Si are the final and initial positions, respectively. The user initially misapplied the integral, treating E as a variable instead of a constant. After correcting the integral, they calculated E as approximately 53851.65 V/m and determined the distance between the two points to be 0.1 m. However, their final calculation of potential difference yielded an incorrect result, prompting further requests for assistance. The discussion highlights the importance of correctly applying integral calculus principles in physics problems.
aliaze1
Messages
173
Reaction score
1

Homework Statement



What is the potential difference between yi= -5cm and yf=5cm in the uniform electric field E = ( 20,000{ i } - 50,000 { j } )V/m?

Homework Equations



∆V = V(sf) - V(si) = -∫ Es ds

with the limits of the integral being sf and si

The Attempt at a Solution



I tried doing the integral, resulting in (E2)/2, and then multiplying proceding as so:

E= √ (200002 + 500002) = 53851.64807
(E2)/2 = 14500000 = x
x |0.05,-0.05 = -{x(0.05) - x(-0.05)} = -{2[x(0.05)]} = -1450000 = incorrect

any help?

thanks!
 
Physics news on Phys.org
I think you're taking the integral as if it were E dE instead of ds (Compare with the case where if it were the integral of x dx, then the integral is x^2/2). E is a constant since it is "uniform", so you can treat it as a constant and pull it out of the integral. If you're only dealing with one dimension (along y), then the integral of E ds simplifies to E(Sf - Si).
 
Anadyne said:
I think you're taking the integral as if it were E dE instead of ds (Compare with the case where if it were the integral of x dx, then the integral is x^2/2). E is a constant since it is "uniform", so you can treat it as a constant and pull it out of the integral. If you're only dealing with one dimension (along y), then the integral of E ds simplifies to E(Sf - Si).

aaahh lol nice...yea that was my mistake...lol whenever i see the integral sign i jump to conclusions without looking at the second part

thanks!
 
well actually...i tried that and it didn't work:

E = 53851.64807

Sf-Si = 0.05 - (-0.05) = 0.1

E * (Sf-Si) = 53851.64807 * 0.1 = 5385.164807 = Incorrect :(

Any help?

Thanks!
 
I have only one attempt left. I noticed that I didn't put the negative sign there, but the computer would tell me to 'check your signs' if this was a sign issue...so i assume it isn't
 
any help?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top