Physics question electric fields

Hi, please could I have some help on question 5bi? I’ve worked out the magnitude but how could the sign be opposite because wouldn’t the electric field then be in the same direction so potential can’t be zero?
Paper: https://pmt.physicsandmathstutor.com/download/Physics/A-level/Past-Papers/CAIE/Paper-4/QP/November%202020%20(v1)%20QP.pdf
Thanks!

the potential due to A is positive as it is a positive charge
the potential due to B is negative so it must be a negative charge
at some point between them the potential is zero, as we can see on the graph

But on another question I was doing, it considered the electrical field due to the charge and if the charges are the opposite, the field is in the same direction so it won’t cancel out?

Oh ok, so would that only apply for field strength?

field strength is a vector quantity
If in one direction it is positive then in the opposite direction it is negative
The field at a point due to several charges can be added but be careful of the sign/direction of each field
Potential is a scalar quantity. It is either positive or negative depending on whether the charge producing it is positive or negative
The potential at a point due to several charges can also be added but be careful of the sign of the individual potentials
Just read the question carefully, as I said they are all different

But if potential is scalar, doesn’t direction not matter?

Hi, please could I have some help on question 5bi? I’ve worked out the magnitude but how could the sign be opposite because wouldn’t the electric field then be in the same direction so potential can’t be zero?
Paper: https://pmt.physicsandmathstutor.com/download/Physics/A-level/Past-Papers/CAIE/Paper-4/QP/November%202020%20(v1)%20QP.pdf
Thanks!

But on another question I was doing, it considered the electrical field due to the charge and if the charges are the opposite, the field is in the same direction so it won’t cancel out?

But if potential is scalar, doesn’t direction not matter?

I am unsure where your understanding of electric potential and field strength goes haywire.
Let's go with the simple electric potential of a point charge.
Electric potential of a point charge Q is given by
$V = \dfrac{k_e Q}{r}$
where ke = 1/(4πϵ₀) = 8.99 × 109 Nm²/C².
Imagine a positive charge of +30 nC is placed fixed at the origin, and we try to calculate the electric potential at a few positions on the positive x-axis.
At x = 0.50 m
$V = \dfrac{8.99 \times 10^9 \times 30 \times 10^{-9} }{0.50} = 540 \text{V}$
At x = 1.0 m
$V = \dfrac{8.99 \times 10^9 \times 30 \times 10^{-9} }{1.0} = 270 \text{V}$
At x = 1.5 m
$V = \dfrac{8.99 \times 10^9 \times 30 \times 10^{-9} }{1.0} = 180 \text{V}$
Next, if we replace the positive charge at the origin with a negative charge of −90 nC, we compute the electric potential at the same positions (0.50 m, 1.0 m, and 1.5 m).
At x = 0.50 m
$V = \dfrac{8.99 \times 10^9 \times (-90 \times 10^{-9}) }{0.50} = -1600 \text{V}$
At x = 1.0 m
$V = \dfrac{8.99 \times 10^9 \times (-90 \times 10^{-9}) }{1.0} = -810 \text{V}$
At x = 1.5 m
$V = \dfrac{8.99 \times 10^9 \times (-90 \times 10^{-9}) }{1.0} = -540 \text{V}$
These calculations show that the positive and negative potential is mainly due to the positive and negative charge, respectively.
Now, let's place the positive charge +30 nC at the origin and the negative charge −90 nC at x=+3.0 m. We calculate the electric potential at some positions.
At x = 0.50 m
$V = \dfrac{8.99 \times 10^9 \times 30 \times 10^{-9} }{0.50} + \dfrac{8.99 \times 10^9 \times (-90 \times 10^{-9}) }{3 - 0.50} = 220 \text{V}$
At x = 0.75 m
$V = \dfrac{8.99 \times 10^9 \times 30 \times 10^{-9} }{0.75} + \dfrac{8.99 \times 10^9 \times (-90 \times 10^{-9}) }{3 - 0.75} = 0 \text{V}$
At x = 1.0 m
$V = \dfrac{8.99 \times 10^9 \times 30 \times 10^{-9} }{1.0} + \dfrac{8.99 \times 10^9 \times (-90 \times 10^{-9}) }{3 - 1.0} = -130 \text{V}$
At x = 1.5 m
$V = \dfrac{8.99 \times 10^9 \times 30 \times 10^{-9} }{1.5} + \dfrac{8.99 \times 10^9 \times (-90 \times 10^{-9}) }{3 - 1.5} = -360 \text{V}$
We can see that the electric potential goes from positive to negative as we move from the positive charge toward negative charge.
In CAIE physics, you need to know how electric potential is connected to electric field:
recall and use the fact that the electric field at a point is equal to the negative of potential gradient at that point
What does the underlined mean is that the electric field points “downhill” in the direction of decreasing potential.
How can we apply this to the current situation (positive charge at x=0 m and negative charge at x=3.0 m)?
Observe the electric potential at x=0.5 m, 0.75 m, 1.0 m and 1.5 m.
As we move in the positive x-direction, the electric potential goes from positive to negative implies that the electric potential is decreasing or going “downhill”, this meant the electric field is pointing in positive x-direction.
This is in agreement with the electric field pattern that you learn at GCSE level as shown below.

From this “the electric field at a point is equal to the negative of potential gradient at that point”, we can find the electric field strength from the electric potential graph Fig. 5.2 given in the question by finding the gradient at different positions.
Say for example the gradient at x = 2.0 cm in Fig. 5.2 is known as the potential gradient, the negative of the potential gradient is the electric field.
Can you see that the electric field at x = 2.0 cm is positive from Fig. 5.2?

THANK YOU SO MUCH! You don't understand how much this has helped. If you don't mind, please could you confirm my summary of what you have said:
So when considering potential, it is based on charges. If there is a positive charge, electrical potential is positive and if there is a negative charge, potential is negative. When moving away from a positive charge towards a negative one, the potential will become more negative.
Electrical field strength is the negative potential gradient. So using a potential graph, you can find the negative gradient to get to electrical field strength. I can see the electric field is positive at x=2cm on Fig 5.2. …

… I have a question regarding where potential and field strength is zero. If you have two equal and opposite charges, the potential will be zero at the centre of both charges. Am I correct to say that if the charges are the same, potential can never be zero? Why is it the case that potential cannot be zero (on a graph there will be no x-intercept) but I don't think that's the explanation.
Also can electric field strength ever be 0?

… Thank you again for taking the time to write it all out, I was getting so muddled between potential and field strength so this helps a lot 🙂

Good summary.
If we have 2 positive charges separated by a distance d, the potential cannot be zero between the 2 charges, but the potential will be zero at infinity.
Why the electric potential cannot be zero at any position between the 2 positive charges?
We can reason from the definition of electric potential and how we compute electric potential.
The “easy” explanation is the electric potential V at a point in space is the sum of the potentials due to each charge and electric potential due to the 2 positive charges is always positive at any positions between the 2 charges. So the algebraic sum of the positive electric potentials cannot be zero.
Can electric field strength ever be zero?
I recommend you try the following simulation to “find out” the answer.
https://physics.bu.edu/~duffy/HTML5/field_potential3.html
You can try out the following cases:
Case 1: left: +30 nC and right: +30 nC
Case 2: left: ‒30 nC and right: ‒30 nC
Case 3: left: +30 nC and right: ‒30 nC
Case 4: left: +30 nC and right: ‒10 nC
Case 5: left: ‒30 nC and right: +10 nC
View both electric potential and electric field on the same graph.
Pay attention to the electric potential when the electric field strength is zero. What can you conclude?
Good that it helps.