A small particle has charge -4.20 and mass 1.10×10−4 . It moves from point , where the electric potential is = 250 , to point , where the electric potential = 740 is greater than the potential at point . The electric force is the only force acting on the particle. The particle has a speed of 5.10 at point .
PLs tell me the answer to dis.
To find the change in electric potential (ΔV) between point A and point B, where the electric potential at point B (VB) is greater than the potential at point A (VA), we can use the formula:
ΔV = VB - VA
Given that VA = 250 and VB = 740, we can substitute these values into the formula:
ΔV = 740 - 250
ΔV = 490
The change in electric potential (ΔV) is 490 units.
Next, let's find the change in kinetic energy (ΔK) using the equation:
ΔK = (1/2)mvf^2 - (1/2)mvi^2
Given that the mass of the particle (m) is 1.10 × 10^(-4) and its initial speed (vi) is 5.10, we can substitute these values into the equation:
ΔK = (1/2)(1.10 × 10^(-4))(5.10)^2 - (1/2)(1.10 × 10^(-4))(0)^2
ΔK = (1/2)(1.10 × 10^(-4))(26.01) - (1/2)(1.10 × 10^(-4))(0)
ΔK = 0.005701
The change in kinetic energy (ΔK) is 0.005701 units.
Since the electric force is the only force acting on the particle, the change in electric potential (ΔV) equals the change in kinetic energy (ΔK). Therefore, we can equate these two values:
ΔV = ΔK
490 = 0.005701
However, the equation above leads to an incorrect solution. There appears to be an error in the given information or calculations. Please check the values again or provide additional information to help resolve the discrepancy.