An electric field `E = (20 hat(i)+30 hat(j))` newton/coulomb exists in the space. If the potential at the origin is taken to be zero, find the potential at `(2m,2m)`.

Explanation

We, have E = `(-dV)/(dr)` so, it can be written in vector form as dV = - `vec(E ).vec(dr)`
note, you can write `vec(E )` as `E_(x) hat(i) + E_(x) hat(j) + E_(z)hat(k) and vec(dr)` = dx`hat(i) + dy hat(j) + dz hat(k)`
therefore `vec(E ).vec(dr) = E_(x). dx + E_(y) . dy + E_(z) .dz`
In the given question the z component of E or the point is not given. so you can write `vec(E ).vec(dr) = E_(x).dx + E_(y).dy`
Now, dV = - `vec(E ).vec(dr)= - (20hat(i) + 30hat(j)).(dx hat(i) + dy hat(j))`
or, dV = - 20 dx - 30 dy
Now we will have to integrate it within limits. Given V = 0 when x = 0 and y = 0 (lower limit and we have to calculate V when x = 2 and y = 2 (upper limit
therefore, `int_(0)^(v) "dv " = -20 int_(x= 0)^(x = 2) "dx" - 30 int_(y-0)^(y =2)` dy
or `[V]_(0)^(v) = - 20 [ X]_(0)^(2) - 30[y]_(0)^(2)`
or, V - 0 = - 20 (2-0) - 30 (2-0)
or , V = -40 - 60 = - 100 Volt.