Comparing Electrostatic Potentials of Positive and Negative Charges

AI Thread Summary
The electrostatic potential of a charge +Q remains constant regardless of the test charge brought into its vicinity, as described by the formula φ(r) = (1/4πε₀)(Q/r). While the electrostatic potential energy differs between a positive test charge (+q) and a negative test charge (-q), with the former having positive energy and the latter negative, the potential itself is unaffected by the sign of the test charge. Thus, the electrostatic potential is the same for both charges at the same distance from +Q. The distinction lies in the potential energy, which is greater for the positive charge. Overall, the key takeaway is that the electrostatic potential is independent of the test charge's sign.
Tony11235
Messages
254
Reaction score
0
This is more of a general question and not a homework question, just to make it clear. Say two test charges are brought separately, one after the other, into the vicinity of a charge +Q. First test charge +q is brought to point B a distance r from +Q. This charge is removed and a test charge -q is brought to the same point. Now do we say that the electrostatic potential of +q is greater because it has a positive sign compared to -q? Or do we say their potentials are the equal because their magnetudes are the same? I assume it's the latter, isn't it?
 
Physics news on Phys.org
The electrostatic potential of the system is always
<br /> \phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}<br />
assuming the test charges are truly infinitesimal (i.e. we neglect their contribution).

The electrostatic potential energy (or just energy) is
<br /> E_{q}= \frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}<br />
in the case of a positive test charge and
<br /> E_{-q}= -\frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}<br />
in the case of a negative test charge. The energy or electrostatic potential energy is different in each case, but the electrostatic potential of the charge Q is the same in both cases. It's mostly a matter of terminology.
 
Physics Monkey said:
The electrostatic potential of the system is always
<br /> \phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}<br />
assuming the test charges are truly infinitesimal (i.e. we neglect their contribution).

The electrostatic potential energy (or just energy) is
<br /> E_{q}= \frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}<br />
in the case of a positive test charge and
<br /> E_{-q}= -\frac{1}{4 \pi \epsilon_0} \frac{qQ}{r}<br />
in the case of a negative test charge. The energy or electrostatic potential energy is different in each case, but the electrostatic potential of the charge Q is the same in both cases. It's mostly a matter of terminology.

So to clear this up, the electrostatic potential energy of the positive test charge is greater?
 
Yes. It has positive energy while the negative charge has negative energy.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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...
Back
Top