Question

the variation of potential with distance r from a fixed point is as shown below. the electric field at r=5m is

1. 2.5 volts/m 2) -2.5 volts/m 3) 2/5 volts/m 4) -2/5 volts/m

Explanation:

Step1: Recall the relation between electric field and potential

The electric field \( E \) is related to the potential \( V \) by the formula \( E = -\frac{dV}{dR} \), where \( \frac{dV}{dR} \) is the slope of the \( V - R \) graph.

Step2: Determine the region for \( R = 5m \)

From the graph, for \( R \) between \( 4m \) and \( 6m \), the potential is decreasing. We need to find the slope of the line in this region.

Step3: Calculate the slope of the \( V - R \) graph for \( 4m \leq R \leq 6m \)

The coordinates of the points on this line: at \( R = 4m \), \( V = 5V \); at \( R = 6m \), \( V = 0V \).
The slope \( \frac{dV}{dR} = \frac{V_2 - V_1}{R_2 - R_1} = \frac{0 - 5}{6 - 4} = \frac{-5}{2} = -2.5 \, V/m \).

Step4: Calculate the electric field

Using \( E = -\frac{dV}{dR} \), substitute the slope:
\( E = -(-2.5) \)? Wait, no. Wait, wait. Wait, the slope here: when \( R \) increases from 4 to 6, \( V \) decreases from 5 to 0. So \( \frac{dV}{dR} = \frac{0 - 5}{6 - 4} = -2.5 \). Then \( E = -\frac{dV}{dR} = -(-2.5) \)? No, wait, no. Wait, the formula is \( E = -\frac{dV}{dR} \). Wait, let's recheck.

Wait, the electric field is the negative of the gradient of the potential. So \( E = -\frac{dV}{dR} \). For the region \( R = 4 \) to \( R = 6 \), the potential is a straight line. Let's take two points: \( (R_1, V_1) = (4, 5) \) and \( (R_2, V_2) = (6, 0) \).

The slope \( \frac{dV}{dR} = \frac{V_2 - V_1}{R_2 - R_1} = \frac{0 - 5}{6 - 4} = \frac{-5}{2} = -2.5 \, V/m \).

Then \( E = -\frac{dV}{dR} = -(-2.5) \)? No, that's wrong. Wait, no. Wait, the formula is \( E = -\frac{dV}{dR} \). Wait, let's think again. The electric field in the radial direction (since \( R \) is distance from a fixed point, assuming spherical symmetry) is \( E = -\frac{dV}{dR} \).

In the region from \( R = 4 \) to \( R = 6 \), the potential is decreasing as \( R \) increases. So the slope \( \frac{dV}{dR} \) is negative (since \( V \) decreases with \( R \)). Then \( E = -\frac{dV}{dR} \) would be positive? Wait, no, maybe I mixed up the direction. Wait, let's take the points: at \( R = 4 \), \( V = 5 \); at \( R = 5 \), what's \( V \)? Let's find the equation of the line from \( (4,5) \) to \( (6,0) \).

The slope \( m = \frac{0 - 5}{6 - 4} = -2.5 \). So the equation is \( V - 5 = -2.5(R - 4) \). So at \( R = 5 \), \( V = 5 - 2.5(1) = 2.5 \, V \). Wait, no, that can't be. Wait, no, when \( R = 4 \), \( V = 5 \); when \( R = 6 \), \( V = 0 \). So the line is \( V = -2.5R + c \). Plugging \( R = 4 \), \( V = 5 \): \( 5 = -2.5(4) + c \) → \( 5 = -10 + c \) → \( c = 15 \). So \( V = -2.5R + 15 \). Then at \( R = 5 \), \( V = -12.5 + 15 = 2.5 \, V \).

Now, the electric field \( E = -\frac{dV}{dR} \). The derivative of \( V \) with respect to \( R \) is \( -2.5 \). So \( E = -(-2.5) = 2.5 \)? No, that's not right. Wait, no, the formula is \( E = -\frac{dV}{dR} \). Wait, maybe I made a mistake in the sign. Let's recall: in the direction of increasing electric field, the potential decreases. So if the potential is decreasing with \( R \), the electric field is in the direction of increasing \( R \), so positive? But let's check the options. The options include -2.5, -2/5, etc. Wait, maybe I messed up the region. Wait, the graph: from R=0 to R=2, it's increasing; R=2 to R=4, constant; R=4 to R=6, decreasing. So at R=5, it's in the decreasing region (R=4 to 6). So the slope of V vs R is (0 - 5)/(6 - 4) = -5/2 = -2.5. Then E = -dV/dR = -(-2.5) = 2.5? But option 2 is -2.5, option 1 is 2.5. Wait, maybe the formula is \( E = \frac{dV}{dR} \) with a sign? Wait, no, the correct formula is \( \vec{E} = - abla V \). In one dimension, \( E = -\frac{dV}{dx} \) (here x is R). So if V is a function of R, then \( E = -\frac{dV}{dR} \).

In the region R=4 to 6, V(R) is a linear function: V(R) = 5 - (5/2)(R - 4) = 5 - 2.5R + 10 = 15 - 2.5R. So dV/dR = -2.5. Then E = -dV/dR = 2.5? But that's option 1. Wait, but let's check the graph again. Wait, the y-axis is potential in volts, x-axis is R in meters. At R=4, V=5; at R=6, V=0. So the slope is (0 - 5)/(6 - 4) = -2.5. So dV/dR = -2.5. Then E = -dV/dR = 2.5 V/m. But option 2 is -2.5. Wait, maybe I got the direction wrong. Wait, maybe the electric field is defined as \( E = \frac{dV}{dR} \) with a negative sign, but maybe the question is considering the magnitude or the sign. Wait, let's check the options. Option 1 is 2.5, option 2 is -2.5. Let's see: when R increases from 4 to 6, V decreases, so the electric field should be in the direction of increasing R (since E points from high potential to low potential). So E is positive in the R direction. But maybe the question is using \( E = \frac{dV}{dR} \) without the negative sign? No, that's incorrect. Wait, maybe I made a mistake in the slope. Wait, let's recalculate the slope. From R=4 to R=6, the change in V is 0 - 5 = -5 V. The change in R is 6 - 4 = 2 m. So slope dV/dR = -5/2 = -2.5 V/m. Then E = -dV/dR = 2.5 V/m. So the answer should be 2.5 volts/m, which is option 1. Wait, but let's check the graph again. Wait, at R=2, V=5; R=4, V=5; R=6, V=0. So from R=4 to R=6, the line goes from (4,5) to (6,0). So the slope is (0 - 5)/(6 - 4) = -2.5. Then E = -dV/dR = 2.5. So option 1.

Wait, but maybe the question is considering the electric field as \( E = \frac{dV}{dR} \), which would be -2.5, but that's incorrect. Wait, no, the correct formula is \( E = -\frac{dV}{dR} \). So if dV/dR is -2.5, then E is 2.5. So the answer is 2.5 volts/m, option 1.

Answer:

1. 2.5 volts/m